Beginning Algebra Parts I and II Workbook

OERWorkbook Algebra Parts I and II 7224 ( .pdf )

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OERWorkbook Algebra Parts I and II 7224word ( .pdf )

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1 Beginning Algebra Parts I and II Palma Benko, David Burger, Yon Kim , Andre a Gloetzer, Michael Walters, Akbar Afsheen, Laura Samuelsen, Deborah Poetsch

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2 Beginning Algebra Part s I & II Contributing Authors Palma Benko, Passaic County Community College David Burger, Passaic County Community College Yon Kim, Passaic County Community College Michael Walters, Passaic County Community College Akbar Afsheen, Bergen County Community College Laura Samuelsen, Hudson County Community College Deborah Poetsch, County College of Morris

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3 Please note this is an uncorrected working draft currently in revision for ADA Compatibility Original Publication Year 2022 Beginning Algebra Workbook OER by Palma Benko, Afsheen Akbar, David Burger, Yon Kim, Deborah Poetsch, Laura Samuelsen, and Michael Walters is licensed under a Creative Commons Attributio n Non Commercial Share Alike 4.0 International License , except where otherwise noted. To learn more about the Open Textbook Collaborative, visit https://middlesexcc.libguides.com/OTCProject Un der this license, any user of this textbook or the textbook contents herein must provide proper attribution as follows: If you redistribute this textbook in a digital or print format (including but not limited to PDF and HTML), then you must retain this attribution statement on your licensing page. If you redistribute part of this textbook, then you must include citation i nformation including the link to the original document and original license on your licensing page. If you use this textbook as a bibliographic reference, please include the link to this work https://opennj. net/l/AA00001559 as your citation. For questions regarding this licensing, please contact library@middlesexcc.edu Funding Statement This material was funded by the Fund for the Improvement of Postsecondary Education (FIPSE) of the U.S. Department of Education for the Open Textbooks Pilot grant awarded to Middlesex College (Edison, NJ) for the Open Textbook Collaborative . Open Textbook Collabora tive The Open Textbook Collaborative with assistance from Brookdale Community College, Ocean County College , Passaic County Community College, and Rowan University . The project engages a consortium of New Jersey community colleges, four year colleges and universities, and workforce partners to develop open educational resources (OER) in career and technical education STEM courses. career pathways in New Jersey’s growth industries including health services, technology, energy, and global manufacturing and supply chain management as identified by the New

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4 Table of Contents Page Chapter 0: Arithmetic 0.1 Integers 6 0.2 Fractions 9 0.3 Order of Operations 1 2 0.4 Properties of Algebra 1 4 Chapter 1: Solving Linear Equations and Inequalities 1.1 One Step Equations 1 7 1.2 Two Step Equations 1 9 1.3 General Linear Equations 2 2 1.4 Solving with Fractions 2 4 1.5 Formulas 2 7 1.8 Application: Number/Geometry 3 1 1.9 Other Applications 3 7 1.10 Solve and Graph Inequalities 4 1 Chapter 2: Graphing 2.1 Graphing: Points and Lines 4 5 2.2 Slope 4 9 2.3 Slope Intercept Form 5 4 2.4 Point Slope Form 5 8 2.5 Parallel & Perpendicular Lines 6 2 Chapter 4: Systems of Equations 4.1 Solving Systems of Equations by Graphing 6 6 4.2 Solving Systems by Substitution 7 2 4.3 Solving Systems by Addition 7 5 4.5 Application: Value Problems 80 4.6 Application: Mixture Problems 8 4 Chapter 5: Polynomials 5.1 Exponent Properties 8 8 5.2 Negative Exponents 9 2 5.3 Scientific Notation 9 4 5.4 Introduction to Polynomials 9 7 5.5 Multiply Polynomials 100 5.6 Multiply Special Products 10 4 5.7 Divide Polynomials 10 8

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5 Chapter 6: Factoring 6.1 FactoringGreatest Common Factor 6.2 FactoringGrouping 6.3: Factoring Trinomials where a=1 6.4 FactoringTrinomials where 6.5 Factoring Special Products 6.6 Factoring Strategies 6.7 Solve by Factoring Chapter 7: Rational Expressions 7.1 Reduce Rational Expressions 7.2 Multiply and Divide Rational Expressions 7.3 Least Common Denominators 7.4 Adding and Subtraction Rational Expressions 7.5 Complex Fractions 7.7: Solving Rational Equations Chapter 8: Radicals 8.1 Square Roots and 8.2 Higher Roots 8.3 Adding Radicals 8.4 Multiplying Radicals 8.5 Dividing Radicals, Rationalizing Denominator Chapter 9: Quadratics 9.1 Quadratics Solving with Radicals 9.2 Solving with Exponents using the Square Root Property 9.3 Completing the Square 9.4 Quadratics Quadratic Formula Answer Key References 113 115 117 119 121 124 126 128 130 133 136 139 142 146 148 150 152 155 159 161 164 166 181

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6 0.1 Integers Learning Objectives : In this section, you will: Add, Subtract, Multiply and Divide Integers Counting Numbers or Natural Numbers : 1,2,3,4. Whole Numbers : 0,1,2,3,4. Integers : Whole Numbers and their opposites, meaning they can be both positive and negative. Zero is also an integer. It is the only integer without a sign. 2, 1, 0, 1, 2, 3, A common application of integers is temperature, which can be positive or negative, in both Fahrenheit and Celsius. Example A) Use as integer: The temperature is 20 degrees below 0. If the temperature was 30 degrees above 0, we’d just write 30. Since the temperature is 20 degrees below 0, we write 20. We can visualize negative numbers using a number line. Values increase as you move to the right and decrease to the left. Example B ) Compare integers: Write < or > to compare the numbers: a) 3 __ 5 b) 4 __ 3 c) 2 __ 5 a) On a number line, 3 is to the left of 5, so 3 < 5 b) On the number line, 4 is to the left of 3, so 4 < 3 c) On a number line, 2 is to the right of 5, so 2 > 5 Add and Subtract Integers To add/subtract signed numbers of the same sign (both positive + + or both negative ) : Add the absolute values of the numbers Keep the sign To add/subtract signed numbers of opposite sign (one positive, one negative : + ) : Subtract the smaller absolute value from the larger absolute value Keep the sign of the larger absolute value number Rewrite subtraction as adding the opposite of the second number: – = + ( – ) and – ( – ) = + Example C) Add: – 8 + ( – 5 ) = 13 Since both numbers are negative, we add their absolute values: 8 + 5 = 13 The result will be negative: – 8 + ( – 5 ) = 8 – 5 = – 13 0 1 2 3 4 5 6 7 8 1 2 3 4 5 6 7 8 Negative Positive

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7 Example D) Add: – 4 + 9 The absolute values of the two numbers are 9 and 4. We subtract the smaller from the larger: 9 – 4 = 5 Since 9 had the larger absolute value and is positive, the result will be positive. – 4 + 9 = 5 Example E ) Add: 5 + ( – 8 ) The absolute values of the two numbers are 5 and 8. We subtract the smaller from the larger: 8 – 5 = 3 Since 8 had the larger absolute value and is negative, the result will be negative. 5 + ( – 8 ) = 5 – 8 = – 3 Example F) Subtract: 10 – ( –3) We rewrite the subtraction as adding the opposite: 10 + 3 = 13 Example G ) Simplify: 3 + ( 4 ) – 2 + 6 – ( 5 ) same signs: add, keep sign work left to right : 7 – 2 + 6 – ( 5 ) same signs: add, keep sign 9 + 6 – ( 5 ) opposite signs: subtract, keep sign of larger 3 + 5 opposite signs: subtract, keep sign of larger 2 OR group all + and : + 6 + 5 3 4 2 add all + and add all – + 11 – 9 opposite signs: subtract, keep sign of larger 2 To multiply or divide two integers If the two numbers have different sign , the result will be negative If the two numbers have the same sign , the result will be positive Example H ) Multiply: a) 4 3 b) 5 ( 6 ) c) 7 ( 4 ) a) The factors have different signs, so the result will be negative: 4 3 = 12 b) The factors have different signs, so the result will be negative: 5 ( 6 ) = 30 c) The factors have the same signs, so the result will be positive: 7 ( 4 ) = 28 Example I ) Divide: a) 40 10 b) 8 ( 4 ) c) a) The numbers have different signs, so the result will be negative: 40 10 = 4 b) The numbers have different signs, so the result will be negative: 8 ( 4 ) = 2 c) The numbers have the same signs, so the result will be positive: = 12 ==========================================================================

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8 Worksheet: 0.1 Integers Write an integer for each situation : 1) I withdraw $100 from my account 2) 20 feet above sea level 3) I lost 5 lbs. Write < or > to compare the numbers : 4) 207 __ 198 5) 23 __ 37 6) 2 __ 7 7) 152 __ 130 Add or s ubtract: 8) – 8 + 3 9) – 1 13 10) 8 + ( – 6 ) 11) 120 + ( – 150 ) 12) 6 18 + 3 13) – 10 – 8 – ( 2 ) 14) 10 – ( 4 ) 15) 26 – ( 12) 16) The temperature was 29 degrees at 6 a.m. It went down 40 degrees by 12 noon. However, it increased by 15 degrees by 10 p.m. What was the temperature at 10 p.m.? 17) In Fargo it was 18F, while in Tacoma it was 43F. How much warmer was Tacoma? 18) Darrel ’s account was overdrawn by $120, before he deposited $450. What is his balance now? Multiply or divide : 19) 7 4 20) 5 ( 8 ) 21) 5 ( 3 ) 22) 48 ( 8 ) 23) 24) ==========================================================================

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9 0.2 Fractions Learning Objectives : In this section, you will: Reduce, add, subtract, multiply, and divide with fractions Converting from mixed number to improper fraction 1. Multiply the whole number by the denominator of the fraction to determine how many pieces we have in the whole. 2. Add this to the numerator of the fraction 3. Use this sum as the numerator of the improper fraction. The denominator is the same. Example A ) Convert 5 to an improper fraction. If we had 5 wholes, each divided into 7 pieces, that’d be 5 7 = 35 pieces. Adding that to the additional 2 pieces gives 35+2 = 37 total pieces. The fraction would be Converting from improper fraction to mixed number 1. Divide: numerator denominator 2. The quotient is the whole part of the mixed number . 3. The remainder is the numerator of the mixed number. The denominator is the same. Example B ) Write as a mixed number. Dividing, 476 = 7 remainder 5. So, there are 7 wholes, and 5 remaining pieces, giving the mixed number 7 Equivalent fractions To find equivalent fractions, multiply or divide both the numerator and denominator by the same number. Example C ) Write two fractions equivalent to By multiplying the top and bottom by 3, = = By dividing the top and bottom by 2, = = Multiply fractions To multiply two fractions, you multiply the numerators, and multiply the denominators: = Example D ) Multiply and simplify = = , which we can simplify to by dividing numerator and denominator by 2. Alternatively, we could have noticed that in , the 2 and 8 have a common factor of 2, so we can divide the numerator and denominator by 2, often called “cancelling” the common factor: = = =

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10 To multiply with mixed numbers, it is easiest to first convert the mixed numbers to improper f ractions. Example E) Multiply and simplify 3 4 Converting these to improper fractions first, 3 = and 4 = , so 3 4 = = . Since 5 and 10 have a common factor of 5, we can cancel that factor: Since 3 and 24 have a common factor of 3, we can cancel that factor: = = 16 Divide fractions To divide two fractions, multiply the first number by that reciprocal of the second number (reciprocal: convert number to upside down form) . Example F ) Divide and simplify We find the reciprocal of and change this into a multiplication problem: 5 8 6 5 = 5 6 8 5 = 1 3 4 1 = 3 4 Example G) Divide and simplify 5 1 Rewriting the mixed numbers first as improper fractions, We find the reciprocal of and change this into a multiplication proble m 11 2 4 3 = 11 4 2 3 = 11 2 1 3 = 22 3 = 7 1 3 A dd and s ubtract fractions with l ike d enominators We can only add or subtract fractions with like denominators. To do this, we add or subtract the numerators . The denominator remains the same: + = and = Example H ) Add and simplify + 7 9 + 5 9 = 7 + 5 9 = 12 9 = 1 3 9 = 1 1 3 A dd and s ubtract with u nlike d enominators 1. Find common denominators 2. Change fractions to equivalent forms having common denominators 3. Add or subtract the numerators. The denominator remains the same Example I ) Add and simplify + Since these don’t have the same denominator, we identify the least common multiple of the two denominators, 4, and give both fractions that denominator. Then we add and simplify. + = + = =

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11 Example J) Subtract and simplify The least common multiple of 8 and 12 is 24. We give both fractions this denominator and subtract. = = Example K) Add and simplify 2 + 5 Rewriting the fractional parts with a common denominator of 12: 2 + 5 Adding the whole parts 2 + 5 = 7 . Adding the fractional parts, + = = 1 . Now we combine these: 7 + 1 = 8 ========================================================================== Worksheet: 0.2 Fractions Convert each mixed number to an improper fraction 1) 4 2) 1 Convert each improper fraction to a mixed number 3) 4) Simplify to lowest terms 5) 6) 7) 8) Multiply and simplify 9) 10) 12 11) 12) 8 4 13) One dose of eyedrops is ounce. How many ounces are required for 40 doses? Divide and simplify 14) 15) 18 16) 3 17) 2 4 18) One dose of eyedrops is ounce. How many doses can be administered from 4 ounces? Add or subtract and simplify 19) + 20) 21) + 22) + 23) + 24) 3 2 25) 8 + 6 26) ==========================================================================

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12 0.3 Order of Operations Learning Objectives : In this section, you will: Evaluate expressions using the order of operations Order of Operations When we combine multiple operations, we need to agree on an order to follow, so that if two people calculate 2 + 3 4 they will get the same answer. To remember the order, some people use the mnemonic PEMDAS: IMPORTANT!! Notice that multiplication and division have the SAME precedence, as do addition and subtraction. When you have multiple operations of the same level, you work left to right . Order of Operations Step 1 . Do anything that is inside parentheses Step 2 . Solve anything that contains an exponent (a power – 5 – the 3 is the exponent and it means the base number is to be multiplied by itself that number of times, so 5 = 5 ( 5 ) ( 5 ) = 125 ) Step 3 . Solve any multiplication or division within the problem, moving from left to right Step 4 . Solve any addition or subtraction within the problem, moving from left to right Exampl e A ) Simplify: 3 5 3+ 6 ( 2 ) We begin with the inside of the parenthe s is , with the exponent: 3 5 9 + 32 ( 2 ) Still inside the parenthesis , we do the division: 3 5 9 + ( 16 ) Inside the parenthesis , we add 9 + ( 16 ) = 9 16 = 7 3 5 ( 7 ) Now multiply 5 ( 7 ) = 35 3 + 35 Add 32 Example B) P 2 ( 12 – 8 ) + ( 3 ) + 4 ( 6 ) E 2 ( 4 ) + ( 3 ) + 4 ( 6 ) MD left to right 2 ( 4 ) + ( 27 ) + 4 6 AS left to right 8 27 24 AS left to right 35 24 59 Example C) P 3 + 4 ( 2 – 6 ) ( 2 ) E 3 + 4 ( 4 ) ( 2 ) MD left to right 3 + 4 ( 16 ) ( 2 ) MD 3 + 64 ( 2 ) AS 3 32 35 P: P arentheses E: E xponents and roots MD: M ultiplication and D ivision AS: A ddition and S ubtraction

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13 If the operations to be performed are in fractional form, solve the numerator first, then the denominator, then reduce. Example D) ( ) Exponent in the numerator, divide in denominator ( ) Multiply in the numerator, subtract in denominator ( ) Add the opposite to simplify numerator, denominator is done. Reduce, divide 20 Our Solution Example E) P Start with ( ), inside ( ): exponents first: = = P Continue ( ): = ( ) =( ) = M = ( ) = = Multiply ========================================================================== Worksheet : 0.3 Order of Op erations Simplify : 1) 4 5 6 2) ( 3 ) 4 3) 18 ( 12 3 ) 4) 19 + ( 7 + 4 ) 5) 20 4 ( 3 2 6 ) 6) 3 + 2 ( 6 3 ) 7) ( ) 8) 9) 6 ( 12 15 ) + 2 10) ( ) ( ) – 11) ( 5 + 2 ) 12) 4 13) 5 14) 15) ( 7 ) 16) Jean’s three pea plants measure 6, 5, and 4 inches tall. Find the mean (average) height . (mean = add up all the numbers, then divide by how many numbers there are) ====================================================== ====================

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14 0.4 Properties of Algebra (Simplify, Evaluate, Translate Expressions) Learning Objectives : In this section, you will: Simplify expressions Evaluate expressions Combine like terms Translate algebraic expressions Vocabulary: Algebraic Expression: An expression that contains at least one variable. Example: + 8 4 ( – ) Terms: All the parts of expressions or equations. Term is a number, a variable, or a product or quotient of numbers and variables raised to powers. Example: 2 + 3 18 terms are 2 , 3x, and 18 Variable: A symbol used to represent a quantity that can change. This is usually a letter. Example: In 3 x + 8, x is the variable. Constant: A value that does not change. Example: In 3 x + 8, 8 is the constant. Coefficient: The number that is multiplied by the variable in an algebraic expression. Example: In 3 x + 8, 3 is the coefficient. Example: 2x+3, variable: x, coefficient: 2, constant: 3 Like terms : terms with exactly the same variables that have the same exponents on the variables are like terms. Example: of like terms would be: 3 & 7 8 & 2 Evaluate algebraic expressions : Replace the variables with their numerical values and follow order of operations. Example A) Evaluate ( + 6 ) when = 3 and = 5 Replace p with 3 and q with 5: ( 3 ) ( 5 + 6 ) Evaluate parenthesis , add: ( 3 ) ( 11) Multipl y : 33 Our solution Combine like terms If we have like terms we are allowed to add (or subtract) the numbers in front of the variables (called coefficients) , then keep the variables the same. Example B ) Simplify: 8 3 + 7 2 + 4 – 3 8 3 + 7 2 + 4 3 Combine like terms 8 2 and 3 + 4 and 7 3 6 + + 4 O ur solution

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15 Distributive Property : multiply a sum or difference, multiply each term by ‘a’ a(b + c) = ab + ac Example C ) Simplify: 4 ( 2 7 ) 4 ( 2 7 ) Multiply each term by 4 8 28 O ur Solution Example D ) Simplify: ( 4 5 + 6 ) ( 4 5 + 6 ) 1 ( 4 5 + 6 ) 4 + 5 6 Our Solution Example E ) S implify : 2 ( 5 8 ) 6 ( 4 + 3 ) 2 ( 5 8 ) 6 ( 4 + 3 ) second 10 16 24 18 14 34 Our Solution Translate algebraic e xpressions Operation + Algebraic Expression x + 28 k 12 8 • w (8)(w) 8w 8(w) n 3 n 3 Words or Phases 28 added to x x plus 28 The sum of x and 28 28 more than x 12 subtracted from k (reverse order!) 12 less than k (reverse order!) Take away 12 from k (reverse order!) k minus 12 the difference of k and 12 8 times w w multiplied by 8 The product of 8 and w 8 groups of w n divided by 3 The quotient of n and 3

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16 Example F) Translate: the product of 8 and 5 less than a number product: multiply, place multiplication where ‘and’ is, o ‘8’ is before multiplication, ‘5 less than a number’ is after 5 less than a number: o a number is any variable: ‘x’ o ‘5 less than a number’ subtraction, reverse order: x 5 Translation: the product of 8 and 5 less than a number 8 ( 5 ) ========================================================================== Worksheet: 0.4 Properties of Algebra (Simplify Expressions) Evaluate each expression if = , = , and = . 1. + 10 2 . + 16 3 . Evaluate each expression if = , = , = , and = . 4 . 2 + 5 . ( 3 ) 6 . 2 ( ) Combine like terms: 7) 7 2 8) 10 6 + 1 9) 2 10) 9 1 + + 4 Distribute: 11) 3 ( 8 + 9 ) 12) ( 5 + 9 ) 13) 10 ( 1 + 2 ) 14) 2 ( + 1 ) Simplify: 15) 12 ( 2 1 ) 16) 9 ( + 10 ) + 5 17) 4 7 ( 1 8 ) 18) 3 ( 4 ) – ( 3 ) Translate: w rite an algebraic expression to the given verbal expression. 19. eight less than a number 20. a number increased by seven 21. the quotient of m and n 22. a number squared 23. the sum of 3 times a and b 24. three times the sum of a and b 25. seven more than the cube of a number 26. one half the product of x and y 27. the product of twice a and b 28. twice the product of a and b 29. two less than five times a number 30. twice a number increased by three times the number 31. the cube of a plus b 32. the cube of the sum of a and b ==========================================================================

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17 1.1 Solving Linear Equations One Step Equations Learning Objectives In this section, you will: Solve one step linear equations by balancing using inverse operations. An equation is a statement asserting that tw o algebraic expressions are equal . Solving equations means to get the variable by itself (isolate). Note: The answer should look like (variable) = (some number), where the variable is never negative Solve using addition and subtraction. Example A) Solve: + 16 = 7 Note: Whatever you do to one side, you MUST do to the other side (keep it balanced). When solving equations, eliminate double signs. Solve: + ( 3 ) = 8 Solve using multiplication and division. Example C) Solve: 5 = 60 + 16 = 7 16 16 r = 23 Get the variable by itself. Right now, 16 is being added to it. Undo the addition by subtracting 16 from both sides. Answer. + ( 3 ) = 8 3 = 8 + 3 = + 3 = 11 Undo the subtraction by adding 3 to both sides. Answer. 5 = 60 5 t 5 = 60 5 = 12 Get the variable by itself. Right now, 5 is being multiplied to it. Undo the multiplication by dividing both sides by 5. Answer. Subtraction Property of Equality Subtracting the same value from both sides of the equation. Addition Property of Equality Adding the same value from both sides of the equation. Division Property of Equality Dividing the same value from both sides of the equation.

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18 R emember: Whatever you do to one side, you MUST do to the other side (keep it balanced). Example D) Solve : = 12 = 12 Since 4 is dividing x, multiply both sides by 4 to clear the fraction. ( 4 ) = 12 ( 4 ) = 48 The fours will cancel each other out. simplifies to 1x = 48 Example E) Solve: = 18 = 18 = 18 = 54 2 = 27 To get rid of multiplying a fraction, multiply by the reciprocal. Multiply s traight across. Check solution: v erify that a given value is a solution to an equation. The two sides must balance. Example F ) Verify that x=7 is the solution to the algebraic equation x 5= 2. We replace x with 7 in the equation. 5 = 2 7 5 = 2 2 = 2 So, 7 is the solution to the 5 = 2 ========================================================================== Worksheet: 1.1 Solving Linear Equations One Step Equations. 1) + 9 = 16 2) 14 = + 3 3) 11 = 16 4) 14 = 18 5) 30 = + 20 6) 1 + = 5 7) 7 = 26 8) 13 + = 19 9) 13 = 5 10) 22 = 16 + 11) 340 = 17 12) 4 = 28 13) 9 = 14) = 15) 20 = 160 16) 20 = 80 ======================================================================== Multiplication Property of Equality Multiplying the same value from both sides of the equation.

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19 1.2 Linear Equations Two Steps Equations Learning Objectives : In this section, you will: Solve a two step equation using addition, subtraction, multiplication, and division. Solve a two step equation by combining like terms. Solve real world problems using twostep equations. TWO STEP EQUATIONS: Work Backwards 1) Find the term with the VARIABLE, you want to isolate that term and then the variable. 2) Identify all operations that is happening to the variable . 3) In REVERSE ORDER of PEMDAS (SADMEP), cancel operations. 4) Use the same number and opposite operations to both sides of the equation. Example A) Solve for x : 13 = 5 + 2 Step 1: Find the term with the variable 13 = 5 + Step 2: What is happening to that term? 13 = + 2 5 is being added to 2 x Step 3: Do the opposite operation to both sides of the equation. 13 = 5 + 2 5 5 Subtraction is the opposite operation of addition, so subtract 5 from both sides. Step 4: Find the variable. 8 = 2 8 = x is being multip lied by 2 (remember that any time there is a number followed by a variable in algebra, it means multiply). Step 6: Do the opposite to both sides of the equation. 8 = 2 2 2 Division is the opposite operation of multiplication, so divide both sides by 2. 4 = x = 4. Check: 13 = 5 + 2 ( 4 ) 13 = 5 + 8 Substitute the answer back into the original equation. Since 13 is 13, our answer is correct. Example B) S olve: 5 + 7 = 7 5 + 7 = 7 Start by focusing on the plus 7 ( 5 + 7 ) 7 = 7 7 Subtract 7 from both sides 5 = 0 Now focus on the multiplication by 5 = Divide both sides by 5 = 0 Our Solution!

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20 Example C ) S olve: 4 2 = 10 4 2 = 10 Start by focusing on the positive 4 ( 4 2x ) = 10 Subtract 4 from both sides 2 = 6 N egative (subtraction) stays on the 2 = Divide both sides by 2 = 3 Our Solution! Example D) Real world problem, solve: An emergency plumber charges $65as a call out fee plus an additional $75 per hour. He arrives at a house at 9:30 AM and works to repair a water tank. If the total repair bill is $196.25, at what time was the repair completed? W e collect the information from the text and convert it to an equation. Unknown time taken in hours this will be our ‘ x ’ The bill is made up of two parts: a call out fee + a per hour fee. $65 as a call out fee 65: independent of x Plus an additional $75 per hour +75x Total Bill = 65+75x The total on the bill was $196.25. So , our final equation is: 196.25 = 65+75x 196.25 = 65+75x 65 65 To isolate x , first subtract 65 from both sides 131.25 = 75x Divide both sides by 75 . = x = 1.75 The time taken was one and three quarter hours (1 hr. 45 mi n) Solution: The repair job was completed at 9:30 + 1:45 = 11:15AM ==========================================================================

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21 Worksheet: 1.2 Solving Linear Equations Two Step Equations. 1) 5 + = 4 2) 2 = 2 + 12 3) 102 = 7 + 4 4) 27 = 21 3 5) 8 + 3 = 77 6) 4 = 8 7) 0 = 6 8) 2 + = 4 9) 8 = 6 10) 5 = 1 11) The product of negative 4 and y increased by 11 is equivalent to 5. 12) Eight more than five times a number is negative 62. 13) You bought a magazine for $5 and four erasers. You spent a total of $25. How much did each eraser cost? 14) Jade is stranded downtown with only $10 to get home. Taxis cost $0.75 per mile, but there is an additional $2.35 hire charge. Write a formula and use it to calculate how many miles she can travel with her money. Determine how many miles she can ride. 15) Jasmins Dad is planning a surprise birthday party for her. He will hire a bouncy castle, and will provide party food for all the guests. The bouncy castle c osts $150 dollars for the afternoon, and the food will cost $3.00 per person. Andrew, Jasmins Dad, has a budget of $300 . Write an equation to help him determine the maximum number of guests he can invite. ================================================= =========================

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22 1.3 General Linear Equations Multi Steps Equations Learning Objectives : In this section, you will: Solve a multistep equation by combining like terms Solve a multistep equation by the distributive property Solve real world problems using multi step equations How to solve a linear equation and find the value of the variable (x) : 1) Use Distributive Property to remove any parentheses , if necessary. 2) Combine like terms on each side , if necessary. 3) Move all variables (x terms) to one side of = sign by adding/subtracting. 4) Move all number terms (constants) to other side of = sign by adding/subtracting. 5) Combine like terms on each side . 6) Divide both sides by coefficient of x term to find ‘x’. 7) Check the solution. Example A) Solve: 3x + 9 = 6x – 27 3 + 9 = 6 27 Notice the variable on both sides, 3x is smaller 3 + 9 + = 6 27 + Add 3x to both sides 9 = 9 27 Focus on the subtraction by 27 9 + = 9 27 + Add 27 to both sides 36 = 9 Focus on the multiplication by 9 = Divide both sides by 9 4 = Our Solution Check: 3 ( 4 ) + 9 = 6 ( 4 ) – 27 12 + 9 = 24 – 27 3 = 3 True, so x = 4 is the solution. Example B ) Solve: 4 ( 2 6 ) = 16 4 ( 2 6 ) = 16 Distribute 4 through parenthesis 8 24 = 16 Focus on the subtraction first 8 24 + = 16 + Add 24 to both sides 8 = 40 Now focus on the multiply by 8 = Divide both sides by 8 = 5 Our Solution! Check: 4 ( 2 ( 5 ) – 6 ) = 16 4 ( 10 – 6 ) = 16 4 ( 4 ) = 16 True, so x = 5 is the solution.

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23 An equation that is true for one or more values of the variable (like the ones above) and false for all other values of the variable is a conditional equation . Example C ) Solve: 3 ( 2 5 ) = 6 – 15 3 ( 2 5 ) = 6 15 Distribute 3 through parenthesis 6 15 = 6 15 Notice the variable on both sides 6 15 = 6 15 Subtract 6x from both sides 15 = 15 Variable is gone! True! When you solve an equation and you end with a T rue statement , the solution set will be: Many Solutions or All Real Numbers . This type of equation is called an Identity. Example D ) Solve: 2 ( 3 5 ) 4 = 2 + 7 2 ( 3 5 ) 4 = 2 + 7 Distribute 2 through parenthesis 6 10 4 = 2 + 7 Combine like terms 6 4 2 10 = 2 + 7 Notice the variable is on both sides 2 10 2 = 2 + 7 2 Subtract 2x from both sides 10 7 Variable is gone! False! When you solve an equation and you end with a F alse statement, the solution set will be: No Solutions. This type of equation is called a Contradiction . Type of equation What happens when you solve it? Solution Conditional Equation True for one or more values of the variables and false for all other variable One or more values Identity True for any value of the variable All real numbers Contradiction False for all values of the variable No solution ========================================================================== Worksheet: 1.3 General Linear Equations Solve each equation. Then state whether the equation is a conditional equation, an identity, or a contradiction. 1) 20 7 = 12 + 30 2) 6 + 12 11 = 7 + 9 + 15 3) 9 ( 2 3 ) 8 = 4 + 7 4) 2 ( 8 4 ) = 8 ( 1 ) 5) 2 5 ( 2 4 ) = 33 + 5 6) 12 + 2 ( 5 3 ) = 9 ( 1 ) 2 7) 4 ( 4 ) ( + 7 ) = 5 ( 3 ) 8) 15 + 32 = 2 ( 10 7 ) 5 + 46 9) 11 ( 8 + 5 ) 8 = 2 ( 40 + 25 ) + 5 10) 23 + 19 = 3 ( 5 9 ) + 8 + 6

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24 1.4 Solving with Fractions Learning Objectives : In this section you will: Solve linear equations with rational coefficients by multiplying by the least common denominator to clear the fractions Solve Equations with Fraction Coefficients Solve equations with fractions: multiply each term by Least Common De nominator . This step will change each coefficient to whole number (the equations stay equivalent to each other). This process is called clearing the equation of fractions . Example A ) Solve: + = Solution Step 1. Find the least common denominator of ail the fractions and decimals in the equation. What is the LCD of , , ? 1 12 + 5 6 = 3 4 LCD = 12 Step 2. Multiply both sides of the equation by that LCD. This clears the fractions and decimals. Multiply both sides of the equation by the LCD, 12. 12 1 12 + 5 6 = 12 3 4 Use the Distributive Property. 12 1 ( 12 ) + 12 5 6 = 12 3 4 Simplify — and notice, no more fractions! + 10 = 9 Step 3. Solve using the General Strategy for Solving Linear Equations. To isolate the variable term, subtract 10. Simplify. + 10 10 = 9 10 = 1 Check: 1 12 + 5 6 = 3 4 1 12 ( 1 ) + 5 6 = 3 4 1 12 + 10 12 = 9 12 9 12 = 9 12

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25 Example B ) Solve: = LCD = 12 , multiply each term by 12 ( 12 ) ( 12 ) = ( 12 ) Reduce each 12 with denominators ( 3 ) 3 = ( 6 ) 7 = ( 2 ) 5 Multiply out each term 9 42 = 10 Focus on subtraction by 42 + 42 + 42 Add 42 to both sides 9 = 52 Focus on multiplication by 9 = Divide both sides by 9 = Our solution In the next example, notice that the 2 is not a fraction in the original equation, but to solve it we put the 2 over 1 to make it a fraction. Example C ) Solve: 2 = + L CD = 6, multiply each term by 6 ( 6 ) ( 6 ) 2 = ( 6 ) + ( 6 ) Reduce 6 with each denominator ( 2 ) 2 ( 6 ) 2 = ( 3 ) 3 + ( 1 ) 1 Multiply out each term 4 12 = 9 + 1 Notice variable on both sides 4 12 = 9 + 1 Subtract 42 from both sides 12 = 5 + 1 Focus on addition of 1 12 = 5 + 1 Subtract 1 from both sides 13 = 5 Focus on multiplication of 5 = Divide both sides by 5 = Our Solution

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26 We can use this same process if there are parentheses in the problem. We will first distribute the coefficient in front of the parentheses, then clear the fractions. See here: Example D ) Solve: ( 5 ) = ( 1 ) Solution 1 2 ( 5 ) = 1 4 ( 1 ) Distribute. 1 2 1 2 5 = 1 4 1 4 1 Simplify. 1 2 5 2 = 1 4 1 4 Multiply by the LCD, 4 . 4 1 2 5 2 = 4 1 4 1 4 Distribute. 4 1 2 4 5 2 = 4 1 4 4 1 4 Simplify. 2 10 = 1 Collect the variables to the left 2 10 = 1 Simplify. 10 = 1 Collect the constraints to the right 10 + 10 = 1 + 10 Simplify. = 9 ========================================================================== Worksheet: 1.4 Solving with Fractions Solve each equation . Make sure any fractional solutions are in simplest form. 1) = 2) = + 3) + 1 = 4) 2 + = 5) = 6) ( 1 + ) = 7) = 8) ( + 15 ) = 3 9) + = + 10) = 3 11 ) =

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27 1.5 Formulas Learning Objectives: In this section you will: Solve linear formulas for a specific variable Solve a Formula for a Specific Variable We have all probably worked with some geometric formulas in our study of mathematics. Formulas are used in so many fields, it is important to recognize formulas and be able to manipulate them easily. It is often helpful to solve a formula for a specific variable. If you need to put a formula in a spreadsheet, you must solve it for a specific variable first. We isolate that variable on one side of the equal sign with a coefficient of one and all other variables and constants are on the other side of the equal sign. When solving formulas for a variable we need to focus on the one variable we are trying to solve for, all the others are treated just like numbers. This is shown in the following example. Two parallel problems are shown, the first is a normal one step equation, the second is a formula that we are solving for x . Example A ) Solve for x: 3 = 12 = In both problems, x is multiplied by something = = To isolate the x, we divide by 3 or w. = 4 = Our Solution We use the same process to solve 3 = 12 for x as we use to solve = for x. Because we are solving for a we treat all the other variables the same way we would treat numbers. Thus, to get rid of the multiplication we divided by w. This same idea is seen in the following example. Example B ) Solve for n: + = for n Solving for n, treat all other variables like numbers + = Subtract m from both sides = Our Solution As and are not like terms, they cannot be combined. For this reason, we leave the expression as . This same one step process can be used with grouping symbols.

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28 Example C ) Solve for ‘y’ : 8 + 7 = 15 Solution We will isolate y on one side of the equation. 8 + 7 = 15 Subtract 8 x from both sides to isolate the term with y. 8 + 7 = 15 Simplify. 7 = 15 8 Divide both sides by 7 to make the coefficient of y one. = Simplify. = Geometric formulas often need to be solved for another variable, too. In the next example, we are given the slope intercept equation of a line and asked to solve for ‘ m ’ , the slope. Example D ) Solve for ‘m’: = + = + for Solving for m, focus on addition first = + Subtract b from both sides = is multiplied by x. = Divide both sides by x = Our Solution It is important to note that we know we are done with the problem when the variable we are solving for is isolated or alone on one side of the equation and it does not appear anywhere on the other side of the equation.

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29 Formulas often have fractions in them : f irst, identify the LCD and then multiply each term by the LCD. After we reduce there will be no more fractions in the problem so we can solve it like any general equation from there. The formula = is used to find the volume of a right circular cone when given the radius of the base and the height. In the next example, we will solve this formula for the height. Example E ) Solve for h: = Solution Write the formula. = 1 3 Remove the fraction on the right. 3 = 3 1 3 ND Simplify 3 = 2 3 = We could now use this formula to find the height of a right circular cone when we know the volume and the radius of the base, by using the formula = Example F) = + Solving for s, focus on what is added to the term with s = + Subtract from both sides = is multiplied by = Divide both sides by = Our Solution Again, we cannot reduce the in the numerator and denominator because of the subtraction in the problem. ==========================================================================

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30 Worksheet: 1.5 Formulas Solve each equation for the specified variable. 1) + = for c 2) = for x 3) = + 2 for L 4) + = for x 5) = 6 ( ) for L 6) = + for h 7) = + for T 8) = for r 9) = for w 10) = for h 11) = 16 for v 12) = for k ==========================================================================

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31 1.8 Applications: Number/Geometry Learning Objectives In this section, you will: Solve number problems. Solve basic geometry problems I. Number Problems: Word problems can be tricky. We will focus on some basic number problems, geometry problems, and parts problems. A few important phrases are described below that can give us clues for how to set up a problem. • A number (or unknown, a value, etc. ) often b ecomes our variable: ex. ‘x’ • Is (or other forms of is: was, will be, are, etc. ) often represents equals ‘ = ’ o ‘ A number is 5’ becomes : x =5 Example A) Solve: If 28 less than five times a certain number is 232. What is the number? 5 28 Subtraction is built backwards, multiply the unknown by 5 5 28 = 232 Is translates to equals 5 28 + = 232 + Add 28 to both sides 5 = 260 The variable is multiplied by 5 = Divide both sides by 5 = 52 The number is 52. Example B ) Solve: Fifteen more than three times a number is the same as ten less than six times the number. What is the number 3 + 15 First, addition is built backwards 6 10 Then, subtraction is also built backwards 3 + 15 = 6 10 Is between the parts tells us they must be equal 3 + 15 = 6 10 Subtract 3 so variable is all on one side

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32 15 = 3 10 Now we have a two step equation 15 + = 3 10 + Add 10 to both sides 25 = 3 The variable is multiplied by 3 = Divide both sides by 3 = Our number is Example C ) Solve: A sofa and a love seat together costs $444. The sofa costs double the love seat. How much do they each cost? Love Seat x With no information about the love seat, this is our x Sofa 2 x Sofa is double the love seat, so we multiply by 2 + = 444 Together they cost 444, so we add. ( ) + ( 2 ) = 444 Replace S and L with labeled values 3 = 444 Parentheses are not needed, combine like terms + 2 = Divide both sides by 3 = 148 Our solution for x Love Seat 148 Replace x with 148 in the original list Sofa 2(148) = 296 The love seat costs $148 and the sofa costs $296. ==========================================================================

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33 Worksheet: 1.8 Number Problems 1. When five is added to three more than a certain number, the result is 19. What is the number? 2. If five is subtracted from three times a certain number, the result is 10. What is the number? 3. When 18 is subtracted from six times a certain 4. A certain number added twice to itself equals 96. What is the number? 6. Sixty more than nine time s a number is the same as two less than ten times the number. What is the number? 7. Eleven less than seven times a number is five more than six times the number. Find the number. 8. If Mr. Brown and his son together had S220, and Mr. Brown had 10 times as much as his son, how much money had each? 9. In a room containing 45 students there were twice as many girls as boys. How many of each were there? 10. An electrician cuts a 30 ft piece of wire into two pieces. One piece is 2 ft longer than the other. How long are the pieces? 11. The total cost for tuition plus room and board at State University is S2,584. Tuition costs S704 more than room and board. What is the tuition fee? ==========================================================================

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34 II. Geometry Problems Another example of translating English sentences to mathematical sentences comes from geometry. We will discuss angles of triangles and perimeter problems. Sum of the measures of the angles in a t riangle: The plural of the word vertex is vertices . All triangles have three vertices : A, B, and C . The lengths of the sides are a, b, and c. The triangle is called by it1s vertices: ABC . The three angles of a triangle are related in a special way. The sum of their measures is 180. Note that we read m A as “the measure of angle A.” So in ABC : m A+m B+m C=180 Perimeter of rectangle: The opposite sides of a rectangle are the same length. We refer to one side of the rectangle as the length, L , and its adjacent side as the width, W . The distance around this rectangle is L+W+L+W, or 2L+2W. This is the perimeter, P , of the rectangle. P=2L+2W

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35 Example D ) Solve: The second angle of a triangle is double the first. The third angle is 40 less than the first. Find the three angles. 2 — 40 The third is 40 less than the first + + = 180 All three angles add to 180 ( ) + ( 2 ) + ( 40 ) = 180 Replace F, S, and T with the labeled values. + 2 + 40 = 180 Here the parenthesis is not needed. 4 40 = 180 Combine like terms, + 2 + 4 40 + 40 = 180 + 40 Add 40 to both sides 4 = 220 The variable is multiplied by 4 = Divide both sides by 4 = 55 Our solution for x 55 Replace x with 55 in the original list of angles 2 ( 55 ) = 110 Our angles are 55, 110, and 15 ( 55 ) — 40 = 15 Example E ) Solve: The Perimeter of a rectangular outdoor patio is 54 ft. The length is 3ft greater than the width. What are the dimensions of the patio? Solution The perimeter formula is standard: P = 2L + 2W. We have two unknown quantities, length, and width . However, we can write the length in terms of the width as L = W + 3. Substitute the perimeter value and the expression for length into the formula. It is oft en helpful to make a sketch and label the sides as in Figure 3. Now we can solve for the width and then calculate the length. = 2 + 2 54 = 2 ( + 3 ) + 2 54 = 2 + 6 + 2 54 = 4 + 6 48 = 4 12 = ( 12 + 3 ) = 15 = The dimensions are L = 15 ft and W=12ft .

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36 Example F ) Solve: The perimeter of a rectangle is 44. The length is 5 less than double the width. Find the dimensions. We will make the length x 2 5 Width is five less than two times the length = 2 + 2 ( 44 ) = 2 ( ) + 2 ( 2 5 ) Replace P, L, and W with labeled values 44 = 2 + 4 10 Distribute through parenthesis 44 = 6 10 Combine like terms 2 + 4 44 + 10 = 6 10 + 10 Add 10 to both sides 54 = 6 The variable is multiplied by 6 = Divide both sides by 6 9 = 2 ( 9 ) 5 = 13 The dimensions of the rectangle are 9 by 13. Worksheet : 1.8 Geometry Problems 1) The second angle of a triangle is the same size as the first angle. The third angle is 12 degrees larger than the first angle. How large are the angles? 2) Two angles of a triangle are the same size. The third angle is 12 degrees smaller than the first angle. Find the measure the angles. 3) The third angle of a triangle is the same size as the first. The second angle is 4 times the third. Find the measure of the angles. 4) The second angle of a triangle is t wice as large as the first. The measure of the third angle is 20 degrees greater than the first. How large are the angles? 5) The perimeter of a rectangle is 150 cm. The length is 15 cm greater than the width. Find the dimensions. 6) The Perimeter of a rectangle is 304 cm. The length is 40 cm longer than the width. Find the length and width. 7) The perimeter of a rectangle is 280 meters. The width is 26 meters less than the length. Find the length and width. 8) The perimeter of a college basketball court is 96 meters, and the length is 14 meters more than the width. What are the dimensions? 9) The perimeter of a rectangle is 608 cm. The length is 80 cm longer than the width. Find the length and width. ======================================================================== ==

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37 1.9 Other Applications: Age, Sales Tax, Discount, and Commission Problems Learning Objectives : In this section, you will: Set up a linear equation to solve an age problem Set up a linear equation to solve a Commission problem Set up a linear equation to solve Sales Tax problem Set up a linear equation to solve Discount problems. I. Age Problems An application of linear equations is what are called age problems. When we are solving age problems, we generally will be comparing the age of two people both now and in the future (or past). To help us organize and solve our problem we will fill out a table for each problem. 1. Fill in the now column. The person we know nothing about is x. 2. Fill in the future/past column by adding/subtracting the change to the now column. 3. Make an equation for the relationship in the future. This is independent of the table. 4. Replace variables in equation with information in future cells of table 5. Solve the e quation for x, use the solution to answer the question Example A) Solve: Carmen is 12 years older than David. Five years ago, the sum of their ages was 28. How old are they now? Five years ago is – 5 in the change column. Carmen is 12 years older than David. We don’t know about David so he is x, Carmen then is x+12 S ubtract 5 from now column to get the change + = 28 The sum of their ages will be 29. So, we add C and D ( + 7 ) + ( 5 ) = 28 Replace C and D with the change cells. + 7 + 5 = 28 Remove parenthesis 2 + 2 = 28 Combine like terms + and 7+5 2 + 2 = 28 Subtract 2 from both sides Age Now 5 Carmen David Age Now 5 Carmen + 12 David + 12 + 12 5 David 5

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38 2 = 26 Notice x is multiplied by 2 2 /2 = 26 /2 Divide both sides by 2 = 13 Our solution for x Replace x with 12 to answer the question Carmen is 25 and David is 13 II. Mark up/Discount problems Mark up/Sales tax formula Given the original cost of an item ‘ C ’ , the mark up/sales tax rate ‘ r ’ , the selling price /total cost ‘ S ’ of the item including the mark up/sales tax rate (‘r’: rate is a percentage and should be converted to a decimal) is given by : = + , Example B) Solve sales tax problem : Imagine that our food cost s $65 in a restaurant, and the sales tax is 8%. What is our total cost? When paying for our meal at a restaurant, we do not pay just the price of the food. We also pay a percentage for sales tax. Then we would pay the original C= $65 plus 8% of that $65. The total cost would be : = + C = 65, r = 8% = .08 total cost = food cost + sales tax = 65 + 0.08( 65) = $70.20 Example C ) Solve markup problem : A retailer acquired a laptop for $2,015 and sold it for $3,324.75. What was the percent markup? Since the retailer acquired the laptop before it was sold, the $2,015 price is the original. We can also consider that the retailer wants to make a profit, and this is a mark up problem. We will use the markup formula, S = C + rC, where C = 2015 and S = 3324.75, to find the mark up rate. 3324.75 = 2015 + r 2015 subtract 2015 from both sides 1309.75 = 2015r divide both sides by 2015 0.65 = r Since the mark up rate is a percentage, then we convert r = 0.65 to a percentage. Hence, the mark up rate is 65% Age Now Carmen 13 + 12 = 25 David 13

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39 III. Commission problems Commission is paid to an employee as an incentive to sell more. A commission is generally a percentage of sales. Commission To find your commission ‘C’, we multiply the sale price ‘S’ by the commission rate (‘r’: rate is a percentage and should be converted to a decimal) is given by: = Example D ) Solve: The Grey family's house was sold for $200,000. How much the real estate agend will earn as commission? How much money will the family have after they pay their real estate agent the 5% commission? C = rS = 0.05( 200,000) = $10,000 The real estate agent will get $10,000 $200,000 $10,000 = $190,000 The family will get $190,000 after they pay their real estate agent. Discount formula Given the regular cost of an item ‘R’, the discount rate ‘r’ (convert percent to decimal), the sale price ‘S’ of the item is given by = Example E ) Solve: Sue bought a sweater for $307.70 after a 15% discount. How much was it before the discount? Since we are looking for the price before the discount was taken and before Sue bought it on sale, our unknown is the regular price, R. The price Sue actually paid for the sweater, $307.70, is the sale price, S. Also, since the sweater is on sale, we subtract from the regular price and we will use the discount formula, where R = 307.70 and r = 15% or 0.15. 307.70 = 1 combine like terms 307.70 = 0.85R divide both sides by .85 362 = R Thus, the regular price of the sweater is $362.

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40 Worksheet : 1.9 Age, Sales Tax, Discount, and Commission Problem. 1) A boy is 10 years older than his brother. In 4 years, he will be twice as old as his brother. Find the present age of each. 2) A father is 4 times as old as his son. In 20 years, the father will be twice as old as his son. Find the present age of each. 3) Find a) the sales tax and b) the total cost: Kim bought a winter coat for $250 in St. Louis, where the sales tax rate was 8.2% of the Purchase price. 4) Diego bought a new car for $ 26, 525. He was surprised that the dealer than added $2,387.25. what was the sales tax rate for this purchase? 5) What is the sale tax rate if a $ 7,594 purchase will have $569.55 of sales tax added to it? 6) Bob is a travel agent. He receives 7% commissio n when he books a cruise for a customer. How much commission will receive for booking a $3900 cruise? 7) Fernando receives 18% commission when he makes a computer sale. How much commission will he receive for selling a computer for $ 2,190? 8) Rikki earned $87 c ommission when she sold a $1450 stove. What rate of commission did she get? 9) Homer received $1140 commission when he sold a car for $28,500. What rate of commission did he get? 10) Marta bought a dishwasher that was on sale for $75 off. The original price of the dishwasher was &525. What was the sale price of the dishwasher? 11) Find a) the amount of discount and b) the sale price: Sergio bought a belt that was discounted 40% from an original price of $29. 12) Find 2) the amount of discount and b) the sale price: Oscar bought a barbecue grill that was discounted 65% from an original price of $ 395. ==========================================================================

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41 1.10 Solve and Graph Inequalities In this section, you will: Solve for the solutions to linear i nequalities. G raph and give interval notation for the solutions to linear inequalities When we have an equation such as x = 4 we have a specific value for our variable. With inequalities we will give a range of values for our variable. To do this we will not use equals, but one of the following symbols: > Greater than Greater than or equal to < Less than Less than or equal to It is often useful to draw a picture of the solutions to the inequality on a number line. We will start from the value in the problem and bold the lower part of the number line if the variable i s smaller than the number and bold the upper part of the number line if the variable is larger. The value itself we will mark with brackets, either ) or ( for less than or greater than respectively, and ] or [ for less than or equal to or greater than or equal to respectively. Once t he graph is drawn, we can quickly convert the graph into what is called interval notation . Interval notation gives two numbers, the first is the smallest value, the second is the largest value. If infinity. If we use either positive or negative infinity, we will always use a curved bracket for that value. Example A) Graph the inequality and give the interval notation < 2 Start at 2 and shade bel ow Use ) for less than Our Graph Interval Notation : ( ADA: Number line shaded: 2, ) at 2; Interval notation: ( , 2)

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42 Example B) Graph the inequality and give the interval notation 1 start at 1 and shade above Use [ for greater than or equal Our Graph Interval Notation : [ ADA: Number line shaded: 1 to [ at 1; Interval notation: [ 1, ) Example C) Give the inequality for the graph: ADA: Number line shaded: 3 to ( at 3 Graph starts at 3 and goes up or greater. Curved bracket means just greater than > 3 Our Solution Solving inequalities is very similar to solving equations with one exception: If we multiply or divide by a negative number, the inequality symbol will need to reverse directions . Example D) Solve and give the result in interval notation : 5 2 11 5 2 5 11 5 Subtract 5 from both sides 2 6 Divide both sides by 2 Divide by a negative – flip symbol! 3 Graph, starting at 3, going down with ] for less than or equal to ADA: Number line shaded: 3, ] at 3; Interval notation: ( , 3]

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43 Example E) Solve and give the result in interval notation : 3 ( 2 4 ) + 4 < 4 ( 3 7 ) + 8 Distribute 6 12 + 4 < 12 28 + 8 Combine like terms 10 12 < 12 20 Move variable to one side 10 12 < 12 20 Subtract 10x from both sides 12 < 2 20 Add 20 to both sides 12 + < 2 20 + 8 < 2 Divide both sides by 2 4 < Be careful with graph, x is larger! ADA: Number line shaded: 4 to ( at 1; Interval notation: (4, ) Note: The inequality symbol opens to the variable, this means the variable is greater than 4. So, we must shade above the 4. ==========================================================================

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44 Worksheet : 1.10 Solve and Graph Inequalities Draw a graph for each inequality and give interval notation. 1) > 5 2) > 4 3) 2 4) 1 5) 5 6) 5 < Write an inequality for each graph. ADA: 7/ Number line shaded: 2, ) at 2 ADA: 8/ Number line shaded: 1, ] at 1 ADA: 9/ Number line shaded: 5 to + , [ at 5 Solve each inequality, graph each solution, and give interval notation. 10) 10 11) 2 + 5 10 12) 7 10 3 1 13) 2 ( 8 ) 18 14) 11 < 8 + 2 15) 8 ( 5 ) > 16 16) > 5 + 17) 6 ==========================================================================

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45 2.1: Graphing: Points and Lines Learning Objectives In this section, you will: Plot ordered pairs of numbers using xy coordinates Graph a linear equation by finding and plotting ordered pair solutions I. Plot Ordered Pairs of Numbers Using XY Coordinates Vocabulary: Coordinate plane: The plane formed by two perpendicular lines called the x axis and y axis. Quadrant: The coordinate plane is divided into four regions. Each region is called a quadrant. xaxis: the horizontal number line. yaxis: the vertical number line. Ordered pair: a pair of numbers that represents a unique point in the coordinate plane. The first value is the x coordinate, and the second value is the y coordinate. o coordinate and 3 is the y coordinate Origin: the center of the coordinate plane. It has coordinates (0, 0). It is the point wher e we always start when we are graphing. Quadrants: Points with (+, +) coordinates are in Quadrant I Points with ( , +) coordinates are in Quadrant II Points with ( , ) coordinates are in Quadrant III Points with (+, ) coordinates are in Quadrant IV Example A ) The first point, A is at (3,2) this means x=3 (right 3) and y =2 (up 2). Following these instructions, starting from the origin, we get our point. The second point, B ( 2,1), is left 2 (negative moves backwards), up 1. This is also illustrated on the graph. x axis origin y axis

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46 Worksheet: 2.1 Plot points in the Cartesian Coordinate System Tell what point is located at each ordered pai r. 1. (3,2) _______ 2. ( 2,3) _______ 3. ( 5,5) _______ 4. ( 7,8) _______ 5. ( 4,4) _______ 6. ( 5,0) _______ Write the ordered pair for each given point. 7. E ___________ 8. M ___________ 9. P ___________ 10. G ___________ 11. Q ___________ 12. N ___________ Plot the follo wing points on the coordinate grid. 13. S ( 6,3) 14. T ( 2, 4) 15. U ( 5 , 8 ) 16. V (4.5, 3.2) 17. W (0, 3) 18. X ( , ) Identify the quadrant in which the point is located (ex 1619) . 19. Point B in graph ________ 20. Point E in graph ___________ 21. Point ( 20, 50) ________ 22. Point (3.5, 100) ___________

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47 II. Graph a linear equation by finding and plotting ordered pair solutions The main purpose of graphs is give a picture of the solutions to an equation. We will do this using a table of values. To graph linear equations (using T tables): 1. Make a T table that contains at least 3 ordered pairs a. Choose whatever numbers you want for “x”, but keep it simple b. Substitute a value for “x”, solve for “y” and fill in the table 2. Make the graph a. The straight line shows all possible solutions to the equation b. If points don’t make straight line, double check the calculations in step #1 and the plotting of ordered pairs Example B ) Graph: = 3 – 2 Use = 1 = 3 ( 1 ) 2 = 3 2 = 1 Use = 0 Use = 2 = 3 ( 0 ) 2 = 3 ( 2 ) 2 = 0 2 = 6 2 = 2 = 4 Example C ) Graph: = 2 + 5 Use = 1 = 2 ( 1 ) + 5 = 2 + 5 = 3 Use = 0 Use = 2 = 2 ( 0 ) + 5 = 2 ( 2 ) + 5 = 0 + 5 = 4 + 5 = 5 = 1 T Table x y 0 2 1 1 2 4 T Table x y 0 5 1 3 2 1

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48 Worksheet : 2.1 Graphing Complete the table to find solutions to each linear equation. 1. = 12 + 3 2. 3 + 2 = 6 Graph by plotting points. 3. = 3 4. = 2 – 4 5. y = + 3 6. – = 6 7. 3 – 2 = 6 8. = 5 9. + 3 = 0 ========================================================================== x y (x,y) 0 4 2 x y (x,y) 0 0 2

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49 2.2 Slope Learning Objectives : In this section, you will: Find the slope of a line given a graph or two points Find x and y intercepts Slope I n everyday life a slope is in the pitch of a roof, the incline of a road, and the slant of a ladder leaning on a wall. In math, the slope define s steepness. Slope = distance moved vertically divided by the distance moved horizontally Ea sier to remember: Slope = rise divided by run . We can find slope graphically and we can find the slope of the line given two points on the line I. Finding slope graphically HOW TO: Find the slope of a line from its graph using = = Step 1. Locate two points on the line whose coordinates are integers. Step 2. Starting with the point on the left, sketch a right triangle, going from the first point to the second point. Step 3. Count the rise and the run on the legs of the triangle. Step 4. Take the ratio of rise to run to find the slope : =

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50 Example A ) Find the slope of the line shown. Step 1. Locate two points on the graph whose coordinates are intergers. Mark (0, 3) and (5,1). Step 2. Starting with the point on the left, sketch a right triangle, going from the first point to the second point. Starting at (0, 3), sketch a right triangle to (5,1). Step 3. Count the ris e and the run on the legs of the triangle. Count the rise. Count the run. The rise is 4 (counting form 3 to 1) . The run is 5 (counting from 0 to 5) . Step 4. Take the ratio of rise to run to find the slope. = Use the slop e formula.Substitut e the values of the rise and run. = = The slope of the line is This mean that y increases 4 units as x increases 5 units

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51 Quick Guide to Slope The direction of a line reveals if the slope is negative or positive. A horizontal line has a slope of 0 and a vertical line has an undefined slope. "Uphill" "Downhill" Horizontal Vertical Positive Slope Negative Slope Slope =0 Slope is Undefined II. Finding the slope of the line given to points on the line Sometimes we’ll need to find the slope of a line between two points when we don’t have a graph. T here is a way to find the slope without graphing. We will use two points: ( x1, y1) to identify the first point and ( x2, y2) to identify the second point. Slope of a Line The slope m of the line containing the points ( x1, y1) and ( x2, y2) is given by = = = Example B) Find the slope of the line through (4, –3) and (2, 2). Step 1 : Label your points Let ( x1, y1) be (4, –3) and ( x2, y2) be (2, 2) Step 2 : Use the slope formula and substitute the value = = = = ( ) = Note: If we let ( x1, y1) be (2, 2) and ( x2, y2) be (4, – 3), then we get the same result. = =

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52 III. Find x and y intercepts X and Y intercepts The points where a line crosses the x axis and the y axis are called the intercepts of a line. The x intercept is the point (a,0) where the line crosses the x axis (y coordinate is 0) . The y intercept is the point (0,b) where the line crosses the y axis (x coordinate is 0) . Example C) Find the x and y intercepts on the graph. X int: line crosses x axis: (4, 0) Y int: line crosses y axis: (0, 2) ADA: Line crosses axis at above listed points : (4,0) and (0,2) FIND THE XAND YINTERCEPTS FROM THE EQUATION OF A LINE To find: the x intercept of the line, let y=0 and solve for x . the y intercept of the line, let x=0 and solve for y . Example D) Find the x and y intercepts from the equation: 2x3y = 12 x intercept of the line, let y = 0 : 2x 3(0) = 12 2x = 12 divide by 2 x = 6 so x int . is ( 6,0) y intercept of the line, let x = 0 : 2(0) – 3 y = 12 3y = 12 divide by 3 y = 4 so y int . is (0 , 4) ==========================================================================

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53 Worksheet : 2.2 Slope Find the slope of the lines. 1. ADA: two point s on the line: (2,3) and (7,6) 2. ADA: two point s on the line: ( 0,5 ) and ( 6,1 ) 3. ADA: two point s on the line: ( 0,1 ) and ( 2,0 ) Find the slope of the line through each pair of points. 4. (13,15), (2,10) 5. (9, 6), (2,10) 6. ( 16,2), (15, 10) 7. ( 18, 5), (5,11) Find the xand y intercepts on the graph. 8. ADA: two point s on the line: ( 1,3 ) and ( 3,3 ) 9. ADA: two point s on the line: ( 5,0 ) and ( 0,5 ) Find the xand y intercepts from the equations: 10. – 4 = 20 11. 3 + 5 = 15 12. = 3 12 13. = 14. = 5

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54 2.3 Graphing: Slope Intercept Form Learning Objectives : In this section you will: 1) Give the equation of a line with a known slope and y intercept. A. Slope Intercept Form When graphing a line, one method is to make a table of values. If we can identify some properties of the line, we may be able to make a graph much quicker and easier. One such method is finding the slope and the y intercept of the equ ation. The slope can be represented by ’ m ’ and the y intercept, where it crosses the axis and x = 0, can be represented by (0, b) where ’ b’ is the value where the graph crosses the vertical y axis. Any other point on the line can be represented by (x, y). Using this information, we will look at the slope formula and solve the formula for y. SLOPE INTERCEPT FORM OF AN EQUATION OF A LINE The slope –intercept form of an equation of a line with slope ’ m ’ and y intercept, (0,b) is, = + Example A) Use the graph to find the slope and y intercept form of the line. ADA: two points on the line: ( 0,1 ) and ( 1,3 ) To find the slope of the line, we need to choose two points on the line. We’ll use the points (0,1) and (1,3). Find the rise and run. = = 2 1 = 2 Find the y intercept of the line. The y intercept is the point (0, 1). We found slope m=2 and y intercept (0,1) = 2 + 1 = +

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55 Example B) Identify the slope and y intercept of the line with equation y = + 5 Solution: We compare our equation to the slope –intercept form of the equation. = + = 3 + 5 Identify the slope. = 3 Identify the y intercept. y intercept is (0,5) Example C) Identify the slope and y intercept of the line with equation + 2 = 6 Solution This equation is not in slope –intercept form. In order to compare it to the slope –intercept form we must first solve the equation for y. Example D) Graph the line of the equation = 4 2 using its slope and y intercept. Solution: Step 1. Fine the slope intercept form of the equation. This equation is in slope intercept form. = 4 2 Step 2. Identify the slope and y intercept. Use = + = + = 4 + ( 2 ) = 4 = 2 , ( 0 , 2 ) Solve for y . + 2 = 6 Subtrac t x from each side. 2 = + 6 Divide both sides by 2. 2 2 = Simplify. 2 2 = 2 + 6 2 Simplify. = 1 2 T + 3 Write the slope – intercept form of the equation of the line. = + = 1 2 T + Identify the slope. = 1 2 Identify the y intercept. y intercept is (0,3)

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56 Step 3. Plot the y intercept. Plot (0,2) Step 4. Use the slope formula = to identify the rise and the run. Identify the rise and the run. = 4 = = 4 = 1 Step 5. Starting at the y intercept, count out the rise and run to mark the second point. Start at (0, 2) and count the rise and the run. Up 4, right 1 Step 6. Connect the points with a line. Connect the two points with a line Note: If a line is not in slope intercept form, solve for ’ y ’ to find the slope intercept form and you can graph it using the method above. ==========================================================================

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57 Worksheet : 2.3 Graphing: Slope Intercept Form Use the graph s to find the slope and y intercept of the lines . 1) ADA: two point s on the line: ( 0,4 ) and ( 6,0 ) 2) ADA: two point s on the line: ( 5,0) and ( 0,5 ) 3) ADA: two point s on the line: ( 0,1 ) and ( 5, 2 ) Identify the slope and y intercept of each line. 4) = 53 – 6 5) 4 – 5 = 8 6) = 4 + 9 Write the slopeintercept form of the equation of each line given the slope and the y intercept. 7) Slope = 2, y intercept = 5 8) Slope = 1, y 9) 10) Slope = 13 , y intercept = 0 Graph the line of each equation using its slope and y intercept. 11) = – 1 12) = 3 + 2 13) 4 – 3 = 12 14) – 2 = 0

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58 2.4 Point Slope Form Learning Objectives: In this section you will: Give the equation of a line with a known slope and point We have two options for finding an equation of a line: slope –intercept or point –slope. Write an equation of the line given a point and slope The slope intercept form is the simplest but requir es us to know the y intercept and slope. Sometimes we only know one or more points (that are not the y intercept). In such a case we must use a different formula instead of slope intercept form. The formula of the equation we will use when the y intercept is not given is called a point slope form (as we will be using a random point and a slope). Derivation of the formula T he slope of an equation is ‘ m ’ , and a specific point on the line be (x1, y1), and any other point on the line be (x, y). We can use the slope formula to make a second equation. Recall slope formula: m, ( x1, y1), (x, y) = = P lug in the values ( ) = ( ) Multiply both side by ( 1 ) = ( ) Our solution/formula This is the point slope formula that requires one ordered pair and formula for the equation of a line. We can easily plug in values in this formula. Point Slope Formula = ( ) When using this formula, we need a slope ‘m’ and a point on the line P(x1, y1) . If the slope is not given, then you must find the slope to use the formula.

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59 Example A) Find an Equation of a Line Given the Slope and a Point : Find an equation of a line with slope m= that contains the point (10,3). Write the equation in slope –intercept form. Step 1. Identify the slope The slope is given = 2 / 5 Step 2. Identify the point The point is given. , 10 , 3 rp Step 3. Substitute the values into the point slope form, = ( ) Simplify. = I ( T rF T ) U rF 3 = ( 10 ) 3 = 4 Step 4. Write the equation in slope intercept form. = 2 5 1 Example B ) Find an equation of a horizontal line that contains the point point slope form. Solution : Every horizontal line has slope 0. Since we have a point and slope, we can substitute the slope and point into the point –slope form, = ( ) Identify the slope. = 0 Identify the point. , 1 , 2 Substitute the values into = ( ) = I ( T rF T ) U rF 2 = 0 ( T rF ( rF 1 ) ) Simplify. U rF 2 = 0 ( T + 1 ) U rF 2 = 0 U = 0 It is in y form but could be written Write in slope –intercept form : Write in point slope form: = 0 + 2 2 = 0 It is a horizontal line . Example C ) Find an equation of a line that contains the point (2,3) with an undefined slope. Write the equation in slope –intercept form. Solution : If the slope is undefined then it is a vertical line. For this we can use the equation of vertical line and substitute. = Identify the slope m = undefined Identify the point (2, 3) Substitute the values into x = b x = 2

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60 Find an Equation of the Line Given Two Points S ometimes we are given just two points and no slope. In that case we need the slope to write out the equation of line. Once we find the slope (using the given points), we can use that and one of the given points to find the equation. Since we will know two points, it will make more sense to use the point –slope form. S lope intercept form requires a slope and a point. Let’s take a look at a problem. Example D ) Find an Equation of a Line Given Two Points Find an equation of a line that contains the points (5,4) and (3,6). Write the equation in slope –intercept form. Step 1. Find the slope using the given points. To use the point slope form, we first find the slope. = = 6 4 3 5 = 2 2 = 1 Step 2 . Choose one point. Choose either point. , 5 , 4 rp Step 3 . Substitute the values into the point slope form, = ( ) Simplify. = I ( T rF T ) U rF 4 = 1 ( 5 ) 4 = 1 + 5 Step 4. Write the equation in slope intercept form. = 1 + 9 In summary we can look at the table below to help us remember what formula to use when writing an equation of line. To Write an Equation of a Line If given: Use: Form: Slope ‘m’ and y intercept (0, b) slope – intercept = + Slope ‘m’ and a point : P (x1, y1) point –slope = ( ) Two points : P1 (x1, y1); P2 (x2, y2) Note: find slope ‘m’ first point –slope = ( ) ==========================================================================

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61 Worksheet: 2. 4 Point Slope Form 1) Write the equation of the line through the point (3, 4) with a slope of 2) Write the equation of the line through the point ( 1, 4) with a slope of 3) Write the equation of the line through the point (2, 2) with a slope of 2 4) Write the equation of the line through the point ( 6, 3) with a slope of 0. 5) Write the equation of a vertical line passing through the point ( 6, 3). 6) Find the equation of a line that contains the points (5,4) and (3,6). This time around use the point (3,6) to write t he equation. 7) Find an equation of a line containing the points (3,1) and (5,6). 8) Find an equation of a line containing the points (1,4) and (6,2). 9) Find an equation of a line containing the points ( 5,4) and ( 5,2). 10) Find an equation of a line containing the points ( 2,3) and (5, 3). ==========================================================================

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62 2.5 Parallel and Perpendicular Lines Learning Objectives: In this section you will: Determine whether the lines are parallel or perpendicular Write an equation of a line given a parallel or perpendicular line Parallel lines have the same slope. = Perpendicular lines have opposite (one +, one ) reciprocal (flipped fractions) slopes . = Example A ) Find the slopes and decide whether the lines are parallel or perpendicular. The above graph has two parallel lines. The slope of the top line is down 2, run 3, or . The slope of the bottom line is down 2, run 3 as well, or . The above graph has two perpendicular lines. The slope of the flatter line us up 2, run 3 or . The slope of the steeper line is down 3, run 2 or Example B ) Find the slope of a line perpendicular to 3 4 = 2 3 4 = 2 To find slope we will put equation in slope intercept form 3 4 3 = 2 3 4 = 3 + 2 Put x term first = + Divide each term by 4 = The slope is the coefficient of x = Slope of first lines. Perpendicular lines have opposite reciprocal slopes = Our Solution Example C ) Determine if the given set of Lines are parallel, perpendicular, or neither. = 2 + 6 ; 2 + = 4

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63 To find out if the lines are parallel or perpendicular, we need to look at their slope. Step 1: Identify the slope of line 1. Slope intercept form: t he slope is given (coefficient of ). = 2 + 6 = 2 Step 2: Identify the slope of line 2. The slope of this equation will have to be found. 2 + = 4 2 + = 4 To find the slope, arrange the equation to slope intercept form by solving for y = 2 4 In this form the slope is given (coefficient of = 2 Step 3: C ompare the slopes of two lines. Since both lines have same slopes, we can identify them as parallel lines. Parallel lines HOW TO: Find an equation of a line parallel or perpendicular to a given line. 1. Find the slope of the given line : ‘m’ 2. Find the slope of the parallel or perpendicular line. Remember: parallel has same slope ‘m’, while perpendicular has opposite slope 3. Identify the point. 4. Substitute the values into the point –slope form, = ( ) . 5. Write the equation in slope –intercept form. Example D ) Find an equation of a line parallel to equation in slope –intercept form. Solution: Step 1. Find the slope of the given line. The line is in slope intercept form, y=2x 3 m= 2 Step 2. Find the slope of the parallel line. Parallel lines have the same slope = 2 Step 3. Ident ify the point. The given point is ( 2,1) @f, , A Step 4. Substitute the values into the point slope form, = ( ) Simplify. = I ( T rF T ) U rF 1 = 2 ( ( 2 ) 1 = 2 ( + 2 ) 1 = 2 + 4 Step 5. Find the slope of the given line. The line is in slope intercept form, = 2 3 = 2 + 5

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64 Example E ) Find an equation of a line parallel to point equation in slope –intercept form. Solution: Step 1. Find the slope of the given line. The line is in slope intercept form, = 2 3 m=2 Step 2. Find the slope of the parallel line. The slopes of perpendicular lines are negative reciprocals. = Step 3. Identify the point. The given point is ( 2,1) , , A Step 4. Substitute the values into the point slope form, = ( ) Simplify. = I ( T rF T ) U rF 1 = ( ( 2 ) 1 = ( + 1 ) 1 = 1 2 1 Step 5. Write the equation in slope intercept form. = Notes: Because a horizontal line is perpendicular to a vertical line, we can say that no slope and zero slope are actually perpendicular slopes . ==========================================================================

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65 Worksheet: 2.5 Parallel and Perpendicular Lines Find the slope of a line parallel to each given line. 1. = 2 + 4 2. = + 5 3. = 4 – 5 4. = 5 5. 6 5 = 20 6. 3 + 4 = 8 Find the slope of a line perpendicular to each given line. 7. = 3 8. = 9. 3 = 6 10. = 1 11. 8 3 = 9 12. 3 = 3 13) Determine if the given lines are parallel, perpendicular, or neither. 3 = 2 + 3 2 + 3 = 2 14) Determine if the given lines are parallel, perpendicular, or neither. + 3 = 4 8 + 2 = 2 15) Determine if the given lines are parallel, perpendicular, or neither. 9 = 16 3 16 4 = 12 Find the equation of the line given the following. Write the answer in slope intercept form. 16 = 2 17 = 18 = 1 19) through: ( 4, 2), parallel to = 11 20) through:(4, 3), perpendicular to + = 1 21 + 2 = 4 22) through:(5, 2), perpendicular to = 0 23 = 6 ==========================================================================

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66 4.1 Solving S ystems of E quations by G raphing Learning Objectives: In this section you will: To solve systems of equation by graphing and identifying the point of intersection So far, we have solved linear equations in one variable like 8 = 2 + 20 . When we have several equations, we call these a system of linear equations. To solve for two variables such as x and y we will need two equations. We are looking for a solution i.e the ordered pair that works in both equations. Remember the graph of a linear equation is a line. For a system of two equations, we will graph two lines. B y finding what the lines have in common, we’ll find the solution to t he system . Most linear equations in one variable have one solution, but we saw that some equations, called contradictions, have no solutions and for other equations, called identities, all numbers are solutions. T here are three possible cases, as shown bel ow The line s intersect . The lines are parallel. Both equations give the same line. Intersecting lines have one point in common. There is one solution to this system. Parallel lines have no points in common. There is no solution to this system. Because we have just one line, there are infinitely many solutions. First, we decide whether a given ordered pair is the solution to the systems of linear equation. Example A ) Is (2,1) the solution to the system ? 3 = 5 + = 3 ( 2 , 1 ) Identify x and y from the ordered pair = 2 , = 1 Plug these values into each equation 3 ( 2 ) ( 1 ) = 5 First equation 6 1 = 5 Evaluate 5 = 5 True

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67 ( 2 ) + ( 1 ) = 3 Second equation, evaluate 3 = 3 True As we found a true statement for both equations using (2,1) we know that (2,1) is the solution. The goal of is to find that ordered pair for each given problem. Example B ) Is ( 5, 3) the solution to the system ? 2 + = 7 3 + 2 = 9 ( 5 , 3 ) Identify x and y from the ordered pair = 5 , = 3 Plug these values into each equation 2 ( 5 ) + ( 3 ) = 7 First equation 10 3 = 7 Evaluate 13 = 7 False , ( 5, 3) is not a solution Since in this case the ordered pair is not a solution to the first equation there is no need to check it for the second one. As for the ordered pair to be the solution for the entire system it must work for both given equations . S olving systems with a graph We should graph two lines on the same coordinate plane to see the solutions of both equations. Our solution is a solution for both lines, this would be where the lines intersect. Example C ) Solve the System of Linear Equations by Graphing 2 + = 7 2 = 6 Step 1. Graph the first equation To graph the first line, write the equation in slope intercept form. 2 + = 7 = 2 + 7 = 2 = 7 2 T + U = 7 T rF 2 = 6

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68 Step 2. Graph the second equation on the same rectangular coordinate system To graph the second line, use intercepts. 2 = 6 (0, 3) (6,0) Step 3. Determine whether the lines intersect, are parallel, or are the same line. Look at the graph of the lines. The lines intersect. Step 4. Identify the solution to the system. If the lines intersect, identify the point of intersection. Check to make sure it is a solution to both equations. This is the solution to the system. If the lines are parallel, the system has no solution. If the lines are the same, the system has an infinite number of solutions. Since the lines intersect, find the point of intersection. Check the point in both equations. The lines intersect at (4, 1). 2 + = 7 2 ( 4 ) + ( 1 ) = 7 8 1 = 7 7 = 7 2 = 6 4 2 ( 1 ) = 6 6 = 6 The solution is (4,1) Example D ) Solve the System of Linear Equations by Graphing = 2 + 1 = 4 1 Both equations in this system are in slope intercept form, so we will use their slopes and y intercepts to graph them. Find the slope and y intercept of the first equation. = 2 + 1 = 2 = 1 Find the slope and y intercept of the first equation. = 4 1 = 4 = 1 Graph the two lines.

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69 Determine the point of intersection. The lines intersect at (1, 3). Since both lines intersect at the ordered pair (1,3), it is the solution for the system. S o far, the lines intersected, and the solution was one point. In the next two examples, we’ll look at a system of equations that has no solution and at a system of equations that has an infinite number of solutions. Example E) Solve the System of Linear Equations by Graphing = 4 = + 1 Solution: = 4 Identify slope and y intercept of each equation = + 1 Fi r st: = , = 4 Now we can graph both equations on the same plane Second = , = 1 To graph each equation, we start at the y intercept and use the slope = to get the next point and connect the dots. The two lines do not intersect! They are parallel! If the lines do not intersect, we know that there is no point that works in both equations, there is no solution Note: We also could have noticed that both lines have the same slope. Remembering that parallel lines have the same slope we wo uld have known there was no solution even without having to graph lines.

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70 Example F) Solve the System of Linear Equations by Graphing 2 6 = 12 3 9 = 18 Solution: 2 6 = 12 3 9 = 18 2 6 2 = 12 2 3 9 3 = 18 3 Subtract x terms 6 = 2 + 12 9 = 3 + 18 Put x terms first = + = + Divide by coefficient of y = 2 = 2 Identify the slopes and y intercepts First: = , = 2 Now we can graph both equations t ogether Second: = , = 2 To graph each equation, we start at the y intercept and use the slope = to get the next point and connect the dots. Both equations are the same line! As one line is directly on top of the other line, we can say that the lines “intersect” at all the points! Here we say we have infinite solutions Once we have both equations in the slope intercept form we can see that they both are the same equations. Moreover, by graphing we can see they both lie on top of each other giving us infinite number of solutions. Determine the Number of Solutions of a Li near System : Graph Number of solutions 2 intersecting lines 1 Parallel lines None Same line Infinitely many ==========================================================================

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71 Worksheet: 4.1 Solving S ystems of E quations by G raphing and I dentifying the P oint of I ntersection 1) Is (2, 4) is the solution to the system ? 4 = 4 2 = 12 4 2) Is ( 3, 1) is the solution to the system ? 3 = 10 4 = 15 3 3) Is ( 2, 0) is the solution to the system ? 5 + = 2 3 + 6 = 12 Solve the following systems of linear equations using graphing method : 4) = 2 + 1 = 2 5) = 2 + 2 + 2 = 2 6) 2 + = 1 2 = 3 7) 3 = 1 = 4 8) 2 + = 4 2 = + 4 9) 5 = 4 10 2 = 10

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72 4.2 Solving Systems by Substitution Learning Objectives: In this section, you will: Solve systems of equations using substitution A system of equations has multiple variables (‘x’, ‘y’ or others). If we can get it down to one variable, we can solve the system just like we have solved equations previously. Our goal is to take this: + = + = and turn it into either a single equation with only ‘ x ’ or only ‘ y ’ in it. Note: A, B, C, D, E, and F are just numbers. Steps for substitution method 1. Solve one of the equations for either variable 2. Substitute the expression from Step 1 into the other equation 3. Solve the resulting equation 4. Substitute the solution in Step 3 into either of the original equations to find the other variable 5. Write the solution as an ordered pair 6. Check that the ordered pair is a solution to both original equations Example A ) Solve the system by substitution: 3 + = 3 (1) 2 + 3 = 5 (2) Solution: Since the first equation has y with a coefficient of 1, we will solve the first equation for y = 3 – 3 (3) 2 + 3 = 5 ( 2) We will then substitute what we have for y equation (3) into equation ( 2) and solve 2 + 3 ( 3 3 ) = 5 Substitution 2 9 – 9 = 5 Distribute 7 9 = 5 Collect like terms 7 = 14 Isolate the variable = 2 Divide away the coefficient Now that we know what x is, we can substitute it into equation (3) to get y = 3 ( 2 ) 3 = 6 3 = 3 So, the so lution is (2 , 3). Remember we can always check our solution by plugging it into the original system.

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73 Example B) Solve the system by substitution: + = 4 (1 ) = 2 + 5 ( 2) Solution: Since the second equation (2) is already solved for x , we can go straight to plugging the solved form into equation ( 1) 2 + 5 + = 4 Substitution 3 + 5 = 4 Collect like terms 3 = 9 Isolate the variable = 3 Divide away the coefficient Now that we know what y is, we can substitute it into equation ( 2) to get x = 2 ( 3 ) + 5 = 6 + 5 = 1 So, the solution is (1 ,3). Remember we can always check our solution by plugging it into the original system. Special Cases : It is important to remember that we have three possible solutions for a system of equations: 1. The system has one solution 2. The system has no solution variable from will cancel out, leaving just numerical values the numbers will not equal one another , making a false statement 3. The system has infinitely many solution variable from will cancel out, leaving just numerical values the numbers will be equal one another , making a true statement Example C) Solve the system by substitution: + 3 = 4 ( 1) 2 6 = 3 ( 2) Solution: Since the first equation has x with a coefficient of 1, we will solve the first equation for x = 4 3 ( 3 ) We can now substitute equation ( 3) into equation ( 2) 2 ( 4 3 ) 6 = 3 Substitution 8 + 6 6 = 3 Distribute 8 = 3 Simplify We see the variable y has cancelled, and the statement left is false since the two numbers are not equal. This means the system has no solution .

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74 Worksheet: 4.2 Solving Systems by Substitution Find the solutions to the systems: 1) = + 2 2 + = 4 2) = + 2 = 2 + 2 3) + 4 = 6 8 + = 81 4) + = 1 4 – 3 = 6 5) = 3 + 4 2 + 6 = 8 6) 2 + = 5 8 – 4 = 24

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75 4.3 Solving Systems by Addition Learning Objectives: In this section, you will: Solve systems of equations using the addition/elimination method A system of equations has multiple variables. If we can get it down to one variable, we can solve the system just like we have solved equations previously. Our goal is to take this: + = + = and turn it into either a single equation with only ‘ x ’ or only ‘ y ’ in it. In the addition, or also called the elimination method, our goal is to have one of the variables match in coefficient and opposite sign so that if the equations are added up, the va riable will cancel. Note: A, B, C, D, E, and F are just numbers. Steps for addition method 1. Write both equations in standard form. If any coefficients are fractions, clear them 2. Pick one variable and multiply either/both equations by a constant to make those variable have the same coefficient with opposite signs 3. Add the equations resulting from Step 2 to eliminate one variable 4. Solve for the remaining variable 5. Substitute the solution from Step 4 into one of the original equations to solve for the other variable 6. Write the solution as an ordered pair 7. Check that the ordered pair is a solution to both the original equations Point : To clear fractions, we multiply the equation by the least common denominator. If both equations have fractions, we repeat this process for the other equation. Be sure when clearing that you multiply every term on both the left and right by the least common denominator. E xample A) Solve the system by the addition method: 3 + = 5 (1) 2 = 0 (2) To solve a system of equations by addition, we start with both equations in standard form. Then we decide which variable will be easiest to eliminate. We want to have the co efficients of one variable be opposites, so that we can add the equations together and eliminate that variable. Here it is easy to eliminate ‘ y ’, since it has 1 and 1 as coefficients : 3 + = 5 (1) 2 = 0 (2) Add (1) and (2) The y’s eliminate and we have one equation with one variable , ‘x’. 5 = 5 Divide both sides by 5 = 1 Substitute ‘x’ to the easier equation (2) to find ‘y’: 2 ( 1 ) – = 0 2 – = 0 = 2 So, the solution is (1, 2)

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76 Example B ) Solve the system by the addition method: 2 + = 7 – 2 = 6 Step 1. Write both equations in standard form. If any c oefficients are fractions, clear them. Both equations are in standard form, + = . There are no fractions. 2 + = 7 – 2 = 6 Step 2. Make the coefficients of one variable opposites. Decide which variable you will eliminate. Multiply one or both equations so that the coefficients of that variables are opposites. We can eliminate the y's by multiplying the first equation by 2. Multiply both sides of 2 + = 7 by 2. 2 + = 7 2 = 6 2 ( 2 + ) = 2 ( 7 ) 2 = 6 Step 3. Add the equations resulting from Step 2 to eliminate one variable. We add the x’s, y’s, and constants. 4 + 2 = 14 2 = 6 5 Step 4. Solve for the remaining variable. Solve for x. Step 5. Substitute the solution from Step 4 into one of the original equations. Then solve for the other variable. Substitute = 4 into the second equation, 2 = 6 Then solve for y. 2 = 6 4 2 = 6 2 = 2 Step 6. Write the solution as an ordered pair. Write it as (x,y). ( 4 , Step 7. Check that the ordered pair is a solution to both original equations. Substitute (4, 1) into 2x+y =7 And x 2y =6 Do they make both equations true? Yes! 2 + = 7 2 = 6 2 ( 4 ) + ( 1 ) = 7 4 2 ( 1 ) = 6 7 = 7 6 = 6 The solution is (4,1) .

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77 Example C ) Note: addition and elimination are the same methods . Solve the system by elimination: 4 T rF 3 U = 9 7 T + 2 U = rF 6 Solution In this example, we cannot multiply just one equation by any constant to get opposite coefficients. So, we will strategically multiply both equations by different constants to get the opposites 4 T rF 3 U = 9 7 T + 2 U = rF 6 Both equations are in standard form. To get opposite coefficients of y, we will multiply the first equation by 2 and the second equation by 3 2 ( 4 T rF 3 U ) = 2 ( 9 ) 3 ( 7 + 2 ) = 3 ( 6 ) Simplify 8 T rF 6 U = 18 21 T + 6 U = rF 18 Add the two equations to eliminate y. 8 T rF 6 U = 18 21 + 6 = 18 29 = 0 Solve for x. = 0 7 + 2 = 6 Substitute x = 0 into one of the original equations. 7 ( 0 ) + 2 = 6 Solve for y. 2 = 6 = 3 Write the solution as an ordered pair. The ordered pair is (0, 3) Check that the ordered pair is a solution to both original equations 4 3 = 9 7 + 2 = 6 4 ( 0 ) 3 ( 3 ) = 9 7 ( 0 ) + 2 ( 3 ) = 6 9 = 9 6 = 6 The solution is (0, 3) Special cases: no solution or infinitely many solutions The system has no solution o variable from will cancel out, leaving just numerical values o the numbers will not equal one another , making a false statement The system has infinitely many solution o variable from will cancel out, leaving just numerical values o the numbers will be equal one another , making a true statement Note: Fractional coefficients : clear the fractions first.

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78 Example D) Solve the system by the addition method: 3x + 4y = 12 = 3 Solution 3 + 4 = 12 = 3 3 4 Write the second equation in standard form. 3 + 4 = 123 4 + = 3 Clear the fractions by multiplying the second equation by 4. 3 + 4 = 12 43 4 + = 4( 3) Simplify. 3 + 4 = 12 3 + 4 = 12 To eliminate a variable, we multiply the second equation by 1. Simplify and add. 3 + 4 = 12 3 4 = 12 0 = 0 This is a true statement. The equations are consistent but dependent. Their graphs would be the same line. The system has infinitely many solutions. After we cleared the fractions in the second equation, did you notice that the two equations were the same? That means we have coincident lines. Solve the system by elimination: 3 + 4 = 12 = 3 3 4

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79 Worksheet: 4. 3 Solving Systems by Addition Find the solutions to the systems: 1. 3 2 = 1 + 2 = 5 2. 3 = 16 2 + = 11 3. + 3 = 11 3 = 9 4. 5 3 = 6 4 12 = 4 5. + = 5 2 3 = 24 6. + = 7 1 2 + 4 5 = 5 7. The sum of two numbers is 39. Their difference is 9. Find the numbers. 8. The sum of two numbers is 15. Their difference is 35. Find the numbers.

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80 4.5 Application: Value Problem s Learning Objectives: In this section, you will: Solve value problems by setting up a system of equations One type of problem systems can solve for us is value problems. These are characterized by an amount of an item and the item having a value. Think about three quarters. You have three of th ese items that has a value of $0.25, giving a total amount of $0.75 Point : Multiply how many of an item you have by its worth to get the value. The equations we use relates to the value. It is often helpful to set up a chart. Number Value Total Item 1 Item 2 Total We fill in the information in the chart based on the problem and use it to make a system of equations for us to work with. The total going horizontally is from the number times the value. Keep in mind not all sections will be filled in. Point : The value column is usually empty in the total spot, as it has no meaning to our problems. When it does have a meaning, we are looking at mixture problems. As an example, if we were interested in how many quarters q and dimes d someone has, our base chart could look like this. Number Value Total Dime d 0.10 0.10d Quarter q 0.25 0 .25q Total If we knew the total amount of coins, we could fill in the last entry in the number column. If we need the amount of money, we could fill in the last entry in the total column. The quarter and dime having a combined value of $0.35 will not help our problem of figuring out how many of each coin we have, so we leave the value total, the last entry in the value column, blank. Points : If the problem is about interest, remember that yearly interest is principal times interest rate. Account Principal Rate Interest Account 1 Account 2 Total

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81 Example A) Natasha has a bank full of nickels and dimes. The total value in her bank is $8.10. The number of dimes is 9 less than twice the number of nickels. How many nickels and how many dimes does Natasha have? Solution: We want the number of nickels and dimes Natasha has, so we will use n for nickels and d for dimes. Since we know the values of each of these, we can fill in what we know so far into our chart. Number Value Total Dimes d 0.10 0.10d Nickels n 0.05 0.05n Total 8.10 This is all the information we must fill into the chart. This does give us one equation that relates the total worth of $8.10 to the total value of each coin. 0 . 10 + 0 . 05 = 8 . 10 We know for two unknowns; we need two equations. The other equation comes from the amounts they compared in the problems. They told us the dimes are so many compared to the nickels, so we will translate that statement. = 2 9 We can solve this system by any method we like, but since the second equation is solved for d already, we will use substitution 0 . 10 ( 2 9 ) + 0 . 05 = 8 . 10 Substitution 0 . 2 0 . 9 + 0 . 05 = 8 . 10 Distribute 0 . 25 0 . 9 = 8 . 10 Collect like terms 0 . 25 = 9 Isolate the variable = 36 Divide away the coefficient Now that we know how many nickels Natasha has, we can substitute it to find the number of dimes. = 2 9 = 2 ( 36 ) 9 = 63 This means for our solution; Natasha has 63 dimes and 36 nickels.

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82 Example B ) The box office at a movie theater sold 147 tickets for the evening show, and receipts totaled $1302. How many $11 adults and how many $8 child tickets were sold? Solution: We want the amount of ticket sales for adults and children. We will let a represent adults and c represent children. To fill in our chart, we know the value of each, and we also know total ticket sales and total value. Number Value Total Child c 8 8c Adult a 11 11a Total 147 1302 We should see that the number column and the total value column form equations for us. We know how many to combine to get the totals out, so we can write as follows. + = 147 8 + 11 = 1302 Again, we will use substitution since both coefficients in the first equation are one. We will solve for ‘ c ’: = 147 We can now substitute the expression for c into the second equation to find a. 8 ( 147 ) + 11 = 1302 Substitution 1176 8 + 11 = 1302 Distribute 1176 + 3 = 1302 Collect like terms 3 = 126 Isolate the variable = 42 Divide away the coefficient Now that we know how adult tickets were sold, we can substitute to find the number of child tickets sold. = 147 = 14742 = 105 This means for our solution, there were 42 adult tickets sold and 105 child tickets sold.

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83 Worksheet: 4.5 Application: Value Problem s Solve the value problems : 1) The ticket office at the zoo sold 553 tickets one day. The receipts totaled $3963. How many $9 adult tickets and how many $6 child tickets were sold? 2) Matilda has a handful of quarters and dimes, with a total value of $8.55. The number of quarters is 3 more than twice the number of dimes. How many dimes and how many quarters does she have? 3) Adnan has $40,000 to invest and hopes to earn 7.1% interest per year. He will put some of the money into a stock fund that earns 8% per year and t he rest into bonds that earns 3% per year. How much money should he put into each fund? 4) A cashier has 30 bills, all of which are $10 or $20 bills. The total value of the money is $460. How many of each type of bill does the cashier have? 5) A trust fund worth $25,000 is invested in two different portfolios. This year, one portfolio is expected to earn 5.25% interest and the other is expected to earn 4%. Plans are for the total interest on the fund to be $1150 in one year. How much money should be invested at each rate?

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84 4.6 Application: Mixture Problem s Learning Objectives: In this section , you will: Solve mixture problems by setting up a system of equations One type of problem systems can solve for us is mixture problems. These are characterized by combining amounts of ingredients together to get a well mixed solution out. An example would be pouring several juices together. The result is a mixture whose concentration of any flavor depends on how much of each juice went in and the amount of flavor in each. Point : Multiply the amount of each item by the concentration, or part, that item contain s that we want to measure. This product gives us the total amount present in each item. Amount Part Total Item 1 Item 2 Total We fill in the information in the chart based on the problem and use it to make a system of equations for us to work with. The total column at the end is for our product of amount time’s part and is a measure of how much of what we care about the item contains. Keep in mind not all sections will be filled in. Point : The last entry in the part column will have a value. This is one of the main differences to notice with value problems. Since we have a resulting solution of some sort, we have a concentration, or part, to mark. As an example, if we were mixing 3 gallons of a 10% salt solution with 2 gallons of a 15% salt solution to get a 12% salt solution, our table would look like the following. Amount Concentration Total Item 1 3 0.10 0.3 Item 2 2 0.15 0.3 Total 5 0.12 0.6 We see that the amount column has a total of the number of gallons of our mixture. We see the rows for each item have a total of the amount multiplied by the part. The total column combines the totals from each item, so we know how much salt is in the mixture. Point : We write the part, or concentration, as a decimal. It is a percentage in the problem, but a decimal in calculation.

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85 Example A) John is making a large batch of chili. He needs to get a combined total of 20 pounds between the meat and beans. If John has budgeted himself $3 per pound for his chili, how many pounds of meat and beans should he buy if meat is $5 a pound and beans are $ a pound? Solution: We want the amount, in pounds, of meat and beans John has, so we will use m for meat and b for beans. With this information, and the cost of each, we can fill in our chart. Amount Part Total Meat m 5 5m Beans b 1 b Total 20 3 60 To solve a problem with two unknowns, we need two equations. We look to the totals to make our equations. We see that the amount column can give us one equation, where the total column can give us another. + = 20 Amount of each item 5 + = 60 Mixture total We know for two unknowns; we need two equations. The other equation comes from the amounts they compared in the problems. They told us the dimes are so many compared to the nickels, so we will translate that statement. = 2 9 We can solve this system by any method we like, but since the second equation is solved for d already, we will use substitution 0 . 10 ( 2 9 ) + 0 . 05 = 8 . 10 Substitution 0 . 2 0 . 9 + 0 . 05 = 8 . 10 Distribute 0 . 25 0 . 9 = 8 . 10 Collect like terms 0 . 25 = 9 Isolate the variable = 36 Divide away the coefficient Now that we know how many nickels Natasha has, we can substitute it to find the number of dimes. = 2 9 = 2 ( 36 ) 9 = 63 This means for our solution; Natasha has 63 dimes and 36 nickels.

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86 Example B) A 90% antifreeze solution is to be mixed with a 75% antifreeze solution to get 360 liters of an 85% solution. How many liters of the 90% and how many liters of the 75% solutions wil l be used? Solution: In this problem we are mixing these two solutions together, the 90% and the 75%, in order to make this 85% solution. We do not know the amounts of either solution going in, so we will assign them to x and y . Here is what we know Type Number Concentration Amount 90% x 0.90 0.90x 75% y 0.75 0.75y 85% 360 0.85 306 Notice that the first two rows are the solutions going into the pot, and the third row is the mixture we get out. Just as we have seen before, the columns make our equations. We know the numbers we are discussing and the amounts. This gives us our system. + = 360 0 . 90 + 0 . 75 = 306 Again, we will use substitution since both coefficients in the first equation are one. We will solve for x . = 360 We can now substitute the expression for x into the second equation to find y . 0 . 90 ( 360 ) + 0 . 75 = 306 Substitution 324 0 . 90 + 0 . 75 = 306 Distribute 324 0 . 15 = 306 Collect like terms 0 . 15 = 18 Isolate the variable = 120 Divide away the coefficient Now that we know how much of the 75% solution to use, we can determine the amount of 90% solution. = 360 = 360 120 = 240 This means for our solution; we need to use 120 liters of the 75% solution with 240 liters of the 90% solution.

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87 Worksheet: 4.6 Application: Mixture Problem s Find the solution to the systems : 1) Carson wants to make 20 pounds of trail mix using nuts and chocolate chips. His budget requires that the trail mix costs him $7.60. per pound. Nuts cost $9.00 per pound and chocolate chips cost $2.00 per pound. How many pounds of nuts and how many pounds of chocolate chips should he use? 2) Greta wants to make 5 pounds of a nut mix using peanuts and cashews. Her budget requires the mixture to cost her $6 per pound. Peanuts are $4 per pound and cashews are $9 per pound. How many pounds of peanuts and how many pounds of cashews should she use? 3) Jotham needs 70 liters of a 50% sol ution of an alcohol solution. He has a 30% and an 80% solution available. How many liters of the 30% and how many liters of the 80% solutions should he mix to make the 50% solution? 4) A scientist needs 65 liters of a 15% alcohol solution. She has available a 25% and a 12% solution. How many liters of the 25% and how many liters of the 12% solutions should she mix to make the 15% solution? 5) A 40% antifreeze solution is to be mixed with a 70% antifreeze solution to get 240 liters of a 50% solution. How many liters of the 40% and how many liters of the 70% solutions will be used?

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88 5.1 Exponent Properties Learning Objectives : In this section , you will: Use exponents Use combinations of the rules for exponents Apply the rules for exponents in a geometry application. E xponents : In the expression 42, the number 4 is the base and 2 is the exponent and called an exponential expression. Exponential expression (notation) In the expression am , the exponent ‘m’ tells us how many times we use the base a as a factor. A monomial in one variable is a term of the form , where ‘ a’ is a constant and ‘ m’ is a whole number. Example A) Evaluate: 2= ( 2 ) ( 2 ) ( 2 ) -------3 factors = 4 ( 2 ) = 8 ( 3 )= ( 3 ) ( 3 ) ( 3 ) ( 3 ) = 9 ( 3 ) ( 3 ) = 27 ( 3 ) = 81 (Order of operations, multiply left to right) 2= 1 ( 2 ) ( 2 ) = 4 (Order of operations, exponent first) ( 2 )= ( 2 ) ( 2 ) = 4 ( 2 )= ( 2 ) ( 2 ) = 4 (Order of operations, exponent first ) Product Rule of Exponents: = When multiply like bases, keep the base, and add the powers. Example B ) Simplify : 3 3 3 Same base, add the exponents 2 + 6 + 1 3 Our Solution Example C ) Simplify: 2 5 Multiply 2.5, add exponents on x, y, and z 10 Our Solution

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89 Example D ) Simplify: = 2 2= 2 ( ) ( ) = Quotient Rule of Exponents: = When dividing with like bases, keep the base and subtract the powers. Note: it is always the numerator's power minus the denominator's power, see negative exponent rule later. Example E ) Simplify: Same base, subtract the exponents 7 Our Solution Example F ) Simplify: Subtract exponents on a, b, and c Our Solution Example G ) Simplify: = = 3 = Power of a Power Rule of Exponents: ( )= When taking a monomial to a power, keep the base and multiply the powers . Example H ) Simplify: We can solve these two different ways : ( ) = ( )( ) = if we u se exponent and Product rule (add exponents) OR ( )= quicker if we use Power rule ( multiply exponents ) Example I ) Simplify: ( ) Put the exponent of 4 on each factor, multiplying powers Our Solution Example J) Simplify: ( 4 ) Put the exponent of 3 on each factor, multiplying powers 4 Evaluate 4 64 Our Solution Example K ) Simplify: a) ( 5 ) = 5= 125 b) ( 3 )= ( 3 )= ( 3 ) ( 3 ) ( 3 ) = 27

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90 Power of a Quotient Rule of Exponents: = When taking a fraction to a power, raise numerator and denominator to that power . Example L ) Simplify: = = Raise numerator and denominator to 4th power, multiply exponents. Example M ) Simplify: Put the exponent of 2 on each factor, multiplying powers Our Solution Example N ) Simplify: = = =( )( ) =( ) = Combination of rules: Example O ) Simplify: 7 ( 2 ) Parenthesis are already simplified, next use power rules 7 ( 8 ) Using product rule, add exponents and multiply numbers 56 Our Solution Example P ) Simplify: ( ) Use power rule in denominator Use quotient rule 3 Our Solution ========================================================================== Rule of Exponents Product Rule of Exponents = = Power of a Power Rule of Exponents ( ) = ( ) = =

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91 Worksheet : 5.1 Exponent Properties Simplify each of the following. 1. 2. ( 2 ) ( 4 ) 3. ( 6 ) ( 3 ) 4. 2 2 2 2 5. ( 3 ) ( 2 ) 6. 2 2 7. 8. 9. 10. 11. 12. ( 5 ) 13. ( 6 )( ) 14. ( 7 )( ) 15. ( 4 ) 16. 17. ( ) 18. ( 2 ) 19. 6 3 20. ( 3 ) 21. 22. 23. 24. 25. ( 3 ) ( 3 ) ==========================================================================

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92 5.2 Negative Exponent Learning Objectives : In this section , you will: Use ‘ 0’ as an exponent Use negative numbers as exponents Apply the rules for exponents in a geometry application. There are a few special exponent properties that deal with exponents that are not positive. Zero Power Rule of Exponents: = 1 Any number or expression raised to the zero power will always be 1. Example A F ) Evaluate: A) = 1 B) 6 = 1 C) ( 7 ) = 1 D) 6 = 1 E) – ( 6 ) = 1 F) 2 + 3 = 1 + 1 = 2 Rules of Negative Exponents: = 1 1 = = Negative exponents in the numerator must be moved to the denominator, likewise, negative exponents in the denominator need to be moved to the numerator. When the base with negative exponent moves, the exponent becomes positive. Example G ) Simplify: = 1 4 = 1 4 = 1 16 4 = 4 Example H ) Simplify: = = ( 3 ) ( 2 ) = 6 = 6 **** All final answers should be written with positive powers.**** ==========================================================================

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93 Worksheet : 5.2 Negative Exponent 1. 5 2. ( 5 ) 3. ( 2) 4. ( 3 ) 5. 4 6 6. ( 2) + 2 7. 4 + 6 8. 9. 10. ( 2 ) 11. 5 12. 2 + 3 13. 14. ( 4 ) 15. ==========================================================================

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94 5.3 Scientific Notations Learning Objectives: In this section , you will: Express numbers in scientific notation. Convert numbers in scientific notation to standard notation. Use scientific notation in calculations. A number is written in scientific notation when it is expressed in the form 10, a n is an integer. A number in scientific notation is always written with the decimal point after the first nonzero digit and then multiplied by the appropriate power of 10. To write long numbers, it is typical to use scientific notation, a system based on the powers of 10. 10= 1 10= 10 10 = . 1 10= 100 10 = . 01 10= 1000 in the same way 10 = . 001 10= 10000 10 = . 0001 Converting Decimal to Scientific Notation Example A) W rite 435,000 in scientific notation (larger than 1 numbers) 4.25 Move decimal point after first digit (the number must be between 1 and 10) 4 . 35 10 The exponent is determined by the number of places the decimal is moved: 5 here . We use positive exponents, since we must multiply 4.56 by 10= 100 , 000 to get back the original number. ‘x’ means multiplication. Example B ) W rite .000456 in scientific notation (smaller than 1 numbers) 4.56 Move decimal point after first digit (the number must be between 1 and 10) 4 . 35 10 The exponent is determined by the number of places the decimal is moved: 4 here . We use nega tive exponents, since we must multiply 4. 35 by 10 = . 0001 to get back the original number. ‘x’ means multiplication. Example C) 10,400,000 in scientific notation equals 1 . 04 10 Example D) .00204 in scientific notation equals 2 . 04 10

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95 Converting from Scientific Notation to Standard Notation (Decimal) We can also turn a number notated scientifically into a standard notationdecimal number by reversing this process : Example E ) Convert 8 . 7 10 into a decimal. 8 . 7 10 = 8,700,000,000 C onvert this to decimal by moving the decimal point 9 places to the right (positive exponent) Example F ) Negative exponent means a number that is less than one: Convert 5 . 4 10 into a decimal. 5 . 4 10 = .00000054 C onvert this to decimal by moving the decimal point 7 places to the left (negative exponent means we divide). Example G ) 6 . 3 10 = move the decimal point 4 places to the right = 63000 Example H ) 9 . 32 10 = move the decimal point 3 places to the left = .00932 Operations on scientific numbers: Example I ) Multiply scientific numbers , find result in scientific notation: ( 2 . 1 10 ) ( 3 . 7 10) Deal with numbers and 10’s separately ( 2 . 1 ) ( 3 . 7 ) = 7 . 77 Multiply numbers 10 10= 10 Use product rule on 10’s and add exponents 7 . 77 10 Our Solution Example J) Divide scientific numbers , find result in scientific notation (Note: this is an example when your number is not scientific notation after the division, and you must change it to scientific notation firsts.) . . Deal with number and 10’s separately . . = 0 . 53 Divide numbers 0 . 53 = 5 . 3 10 Change this number into scientific notation = 10 Use product and quotient rule, using 10 from the conversion Be careful with signs: ( 1 ) + ( 3 ) ( 7 ) = ( 1 ) + ( 3 ) + 7 = 3 5 . 3 10 Our Solution

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96 Worksheet : 5.3 Scientific Notations Write in scientific notation: 1. 575 2. 87,400 3. 0.643 4. 0.000802 Write in decimal notation: 5. 2 . 54 10 6. 6 . 19 10 3 7. 4 . 64 10 8. 7 10 Solve the following problems : 9. The distance from the earth to the nearest star outside our solar system is approximately 25,700,000,000,000. When expressed in scientific notation, what is the value of n. 2 . 57 10 10. One light year is approximately 5 . 87 10 miles. Use scientific notation to express this distance in feet (Hint: 5,280 feet = 1 mile). 11. John travels regularly for his job. In the past five years he has traveled approximately 355,000 miles. Convert his total miles into scientific notation. Multiply/divide scientific numbers, find result in scientific notation: 12. ( 7 10 ) ( 2 10 ) 13. ( 2 . 6 10 ) ( 6 10 ) 14. . . 15. . .

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97 5.4 Introduction to Polynomials Learning Objectives : In this section , you will: Identify polynomials, monomials, binomials, and trinomials Add and subtract monomials Add and subtract polynomials Evacuate a polynomial for a given value. Polynomial —A monomial, or two or more monomials combined by addition or subtraction. monomial —A polynomial with exactly one term is called a monomial. binomial —A polynomial with exactly two terms is called a binomial. trinomial —A polynomial with exactly three terms is called a trinomial. Example A) Some examples of polynomials: Polynomial + 1 4 7 + 2 4 + + 8 9 + 1 Monomial 14 8 9 13 + 7 4 16 3 9 7 + 12 9 + 2 8 6 + 8 + 3 1 The Degree of a Polynomial The degree of a term is the sum of the exponents of its variables. The degree of a constant is 0. The degree of a polynomial is the highest degree of all its terms. Example B ) Some examples of finding number of terms and degrees: Polynomial Number of terms Type Degree of terms Degree of polynomial 7 5 + 3 3 Trinomial 2,1,0 2 2 3 4 6 + 8 5 Polynomial 5,3,2,1,0 5

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98 Evaluate polynomials : replace the variable with the given number value and evaluate the polynomial. Example C ) Evaluate the trinomial: 2 4 + 6 when x = 4 2 4 + 6 w hen x = 4 Replace variable x with 4 2 ( 4 ) 4 ( 4 ) + 6 Exponents first 2 ( 16 ) 4 ( 4 ) + 6 Multiplication (we can do all terms at once) 32 + 16 + 6 Add 54 Our Solution Add/subtract polynomials: we combine like terms (add/subtract coefficients and keep variable) Example D ) Add 4 2 + 8 and 3 9 – 11 ( 4 2 + 8 ) + ( 3 9 – 11) Combine like terms 4 + 3 and 8 11 7 9 2 3 Our Solution Example E ) Subtract 3 + 6 4 from 5 2 + 7 Note: remember; ‘subtract from’ will reverse order ( 5 2 + 7 ) ( 3 + 6 4 ) Distribute negative through second part 5 2 + 7 3 6 + 4 Combine like terms 5 3 , 2 6 , and 7 + 4 2 8 + 11 Our solution Example F ) Simplify: ( 2x – 5y ) – ( 3x + 2y ) ( 2x – 5y ) – ( 3x + 2y ) Distribute negative through second part 2x – 5y – 3x 2y Combine like terms 2x 3x – 5y 2y Combine like terms (2 3)x +( 5 2)y x 7y Our s olution ==========================================================================

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99 Worksheet : 5.4 Introductions to P olynomial s 1) Evaluate: For the polynomial 5 8 + 4 , find the value when: a) x = 4 b) c) x = 0 2) Determine whether each polynomial is a monomial, binomial, trinomial, or other polynomial. Then, find the degree of each polynomial. a) b) 8 7 3 c) 3 5 + 9 d) 81 4 e) 3 3) Add or subtract: a) 25+ 15 b) 16 ( 7 ) 4) Find the sum: ( 7 2 + 9 ) + ( 4 8 7 ) 5) Add or subtract: a) ( 7 + 8 ) + ( 3 + 2 ) b) ( ) ( + 3 4 ) c) ( 3 ) + ( 2 + 4 ) ( 3 + ) 6) Subtract: ( 4 6 3 ) ( 2 + 7 ) 7) Subtract ( 9 + 2 ) from ( 12 + 6 ) 8) Find the diffrence of ( 3 18 ) and ( + 5 20) =========================================================================

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100 5.5 Multiply Polynomials Learning Objectives : In this section , you will: Multiply a monomial and a polynomial Multiply two polynomials Multiply binomials by the Foil method Multiply a polynomial by a monomial : use the distributive property multiply coefficients (numbers in front of variables) add exponents of like variables Example A) Multiply: 2 ( 4 + 3 5 ) Distribute. 2 4 + ( 2 ) 3 ( 2 ) 5 ) Multiply. 8 6 + 10 Multiply a b inomial by a b inomial : Two methods: distribute or FOIL Example B ) Multiply using the distributive property and FOIL Method: ( + 3 ) ( + 5 ) FOIL stand for ‘First, Outer, Inner, Last’ Distributive Property FOIL ( + 3 ) ( + 7 ) ( + 7 ) + 3 ( + 7 ) + 7 + 3 + 21 F O I L + 7 + 3 + 21 F O I L + 10 + 21 + 10 + 21

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101 Example C ) Multiply using distributive property : ( 4 + 3 ) ( 2 – 5 ) Distribute. 4 ( 2 5 ) + 3 ( 2 5 ) Distribute again. 8 20 + 6 15 Combine like terms. 8 14 15 HOW TO Multiply two binomials using the FOIL method Step 1. Multiply the First terms. Step 2. Multiply the Outer terms. Step 3. Multiply the Inner terms. Step 4. Multiply the Last terms. Step 5. Combine like terms, when possible. Example D ) Multiply using FOIL method: ( + 4 ) ( 1 ) Multiply the First. + _ _ + _ _ + _ _ F O I L Multiply the Outer. + _ _ + _ _ F O I L Multiply the Inner. + 4 + _ _ F O I L Multiply the Last. + 4 4 F O I L Combine like terms – there are none. + 4 4

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102 Multiply a polynomial by a polynomial : use distributive property Example E ) Multiply : ( + 3 ) ( 2 – 5 + 8 ) Distribute. ( 2 – 5 + 8 ) + 3 ( 2 5 + 8 ) Multiply. 2 5 + 8 + 6 15 + 24 Combine like terms 2 + 7 + 24 Example F ) Multiply : ( 2 5 ) ( 4 7 + 3 ) Distribute 2x and 5 ( 2 ) ( 4 ) + ( 2 ) ( 7 ) + ( 2 ) ( 3 ) 5 ( 4 ) 5 ( 7 ) 5 ( 3 ) Multiply out each term 8 14 + 6 20 + 35 15 Combine like terms 8 34 + 41 15 Our Solution Example G ) Us e three methods to multiply binomial s : ( 4 5 ) ( 2 ) Distribute FOIL Rows 4 ( 2 ) 5 ( 2 ) 8 4 10 5 8 14 5 2 ( 4 ) + 2 ( 5 ) ( 4 ) ( 5 ) 8 10 4 + 5 8 14 + 5 2 4 5 10 + 5 8 4 8 14 TU + 5 U ==========================================================================

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103 Worksheet : 5.5 Multiply Polynomials Find each product: 1. 6 ( 7 ) 2. 4 ( 8 + 4 ) 3. 3 ( 6 + 7 ) 4. ( + 6 ) ( + 8 ) 5. ( + 3 ) ( 5 ) 6. ( 8 ) ( 4 + 8 ) 7. ( 7 6 ) ( 7 + 6 ) 8. ( 5 + ) ( 5 2 ) 9. ( 7 ) ( 6 + 5 ) 10. ( 6 4 ) ( 2 2 + 5 ) 11. 3 ( 3 4 ) ( 2 + 1 ) 12. 7 ( 5 ) ( 2 ) 13. ( 2 ) ( + 3 ) ( 4 ) 14. Find the formula s for the perimeter and area of a r ectangle where the = 2 1 and the = 3 + 2 15. Find the formula s for the perimeter and area of a rectangle in terms of ‘w’ where the length is 2 more than the width (‘w’). 16. Find the formula s for the perimeter and area of a rectangle in terms of ‘w’ where the length is 3 less than twice the width (‘w’). ==========================================================================

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104 5.6 Multiply Special Products Learning Objectives : In this section , you will: Square binomials. Find the product of the sum and difference of two terms. Find greater powers of binomials. Squaring a Binomial Let's start by looking at ( + 9 ) What does this mean? ( + 9 ) It means to multiply ( + 9 ) by itself. ( + 9 ) ( + 9 ) Then, using FOIL, we get: + 9 + 9 + 81 Combining like terms gives: + 18 + 81 Square of a Binomial The square of a binomial is a trinomial consisting of the square of the first term + twice the product of the two terms + the square of the last term. For x and y, the following hold. ( + ) = + 2 + ( = T rF 2 TU + U Example A) Square the binomial ( + ) using distributive property ( + ) Squared is same as multiplying by itself ( + ) ( + ) Distribute ( + ) ( + ) + ( + ) Distribute again through final parenthesis + + + Combine like terms + + 2 + Our Solution

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105 Example B ) Square the binomial ( x 5)2 using perfect square formula above ( 5 ) Recognize perfect square Square the first 2 ( ) ( 5 ) = 10 Twice the product ( 5 )= 25 Square the last 10 + 25 Our Solution Example C ) Square the binomial ( 2 + 5 ) using perfect square formula ( 2 + 5 ) Recognize perfect square ( 2 ) = 4 Square the first 2 ( 2 ) ( 5 ) = 20 Twice the product 5 = 25 Square the last 4 + 20 + 25 Our Solution A conjugate pair is two binomials of the form (a – b) and (a + b) Product of conjugates (or difference of squares) The product is called a difference of squares. To multiply conjugates, square the first term, square the last term, write it as a difference of squares. Example D ) Multiply ( + ) ( – ) ( + ) ( ) Distribute ( + ) ( + ) ( + ) Distribute a and b + Combine like terms Our Solution

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106 Example E ) Multiply ( 5 ) ( + 5 ) ( 5 ) ( + 5 ) R ecognize sum and difference 25 Square both, put subtraction between. Our Solution Example F ) Multiply ( 2 – 6 ) ( 2 + 6 ) ( 2 – 6 ) ( 2 + 6 ) Recognize sum and difference 4 36 Square both, put subtraction between. Our Solution Example G ) Review the difference between the three problems : ( 4 7 ) ( 4 + 7 ) 16 49 ( 4 + 7 ) 16+ 56 + 49 ( 4 7 ) 16 56 + 49 Binomial Squares Product of Conjugates ( + ) = + 2 + ( — ) ( + ) = ( — ) = 2 + Squaring a binomial Multiplying conjugates Product is a trinomial . Product is a binomial . Inner and outer terms with FOIL are the same . Inner and outer terms with FOIL are opposites. Middle term is double the product of the terms There is no middle term. ==========================================================================

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107 Worksheet: 5.6 Multiply Special Products Find each product: 1) ( + 5 ) 2) ( 2 1 ) 3) ( 3 2 ) 4) ( 3 ) 5) + 6) 3 ( 4 ) 7) 5 ( ) 8) ( 3 ) ( + 3 ) 9) ( + 7 ) ( 7 ) 10) ( 3 1 ) ( 3 + 1 ) 11) ( 5 2 ) ( 5 + 2 ) 12) ( + 11) ( 11) 13) ( ) ( + ) 14) 5 ( + 1 ) ( 1 ) 15) 3 ( 2 + 1 ) ( 2 1 ) 16) ( 5 ) 17) ( 2 3 ) ==========================================================================

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108 5.7 Divid e Polynomials Learning Objectives : In this section , you will: Divide a polynomial by a monomial. Divide a polynomial by a polynomial. Apply polynomial division in a geometry application. Review: Divide Monomials : W e’ll try to re discover the property by looking at some examples. Consider And What do they mean? Use the Equivalent Fractions Property. Simplify. 1 QUOTIENT PROPERTY FOR EXPONENTS If a is a real number, a 0, and m and n are whole numbers, then = , > = 1 , > Example A) Divide or find the quotient of 56 and 8 56 8 Rewrite as a fraction. 56 8 Use fraction multiplication. 56 8 Simplify and use the Quotient Property. 7 Example B ) Divide or find the quotient of 8 and 10 Rewrite as a fraction Use fraction multiplication Simplify

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109 Divide a polynomial by a monomial : divide each term in the numerator by the monomial in denominator. Example C ) Find the quotient: ( 15 35 ) ( 5 ) ( 15 35 ) ( 5 ) Rewrite as a fraction. 15 35 5 Separate the terms 15 5 35 5 Simplify 3 + 7 Example D ) Divide: Divide each term in the numerator by 3 + Reduce each fraction, subtracting exponents 3 + 2 6 8 Our Solution Example E ) Divide: Divide each tern in the numerator by 4x^2 + + Reduce each fraction, subtracting exponents Remember negative exponents are moved to denominator 2 + 1 + Our Solution

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110 Divide a polynomial by a polynomial . To divide a polynomial by a polynomial , we follow a procedure similar to long division of numbers. Example F ) Divide (x2 + 9x + 20) by (x + 5) Note: realize that we cannot follow processes earlier , since divisor is not monomial . ( 8 + 9 + 20 ) ( + 5 ) Write it as a long division problem. Be sure the dividend is in standard form. + 5 + 9 + 20 Divide by x. It may help to ask yourself, “What do I need to multiply x by to get ? " x + 5 + 9 + 20 Put the answer, x, in the quotient over the x term. Multiply x times + 5 . Line up the like terms under the dividend. x + 5 + 9 + 20 + 5 Subtract + 5 + 9 . You may find it easier to change the signs and then add. Then bring down the last term, 20. x + 5 + 9 + 20 + 5 4 + 20 Divide 4x by x. It may help to ask yourself, “What do I need to multiply x by to get 4x?” Put the answer, 4, in the quotient over the constant term. x + 5 + 9 + 20 + 5 4 + 20 4 + 20 Multiply 4 times + 5 x + 4 + 5 + 9 + 20 + 5 4 + 20 4 + 20 Subtract 4 + 20 from 4 + 20 x + 4 + 5 + 9 + 20 + 5 4 + 20 4 + 20 0 Check : Multiply the quotient by the divisor. You should get the divided. ( + 4 ) ( + 5 ) + 9 + 20

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111 Steps of Dividing Polynomials : 1. Divide front terms 2. Multiply this term by the divisor 3. Change the sign of the terms and combine 4. Bring down the next term 5. Repeat Example G ) Find the quotient: ( – + 5 – 6 ) ( + 2 ) ( – + 5 – 6 ) ( + 2 ) Write it as a long division problem. Be sure the dividend is in standard form with placeholders for missing terms. + 2 + 0 – + 5 – 6 Divide , in the quotient over the term. Multiply times + 2 . Line up the like terms Subtract and then bring down the next term. + 2 + 0 – + 5 – 6 ( + 0 ) 2 4 T Divide rF 2 T by x. Put the answer, 2 , in the quotient over the term. Multiply 2 times x+1. Line up the like terms Subtract and bring down the next term. 2 + 2 + 0 – + 5 – 6 ( + 0 ) 2 ( 2 4 ) 3 + 5 T Divide 3 T by x. Put the answer, 3x, in the quotient over the x term. Multiply 3x ties x+1. Line up the like terms Subtract and bring down the next term. 2 + 3 + 2 + 0 – + 5 – 6 ( + 0 ) 2 ( 2 4 ) 3 + 5 ( 3 + 6 ) 6 Divide 2 + 3 + 2 + 0 – + 5 – 6 ( + 0 ) 2 ( 2 4 ) 3 + 5 ( 3 + 6 ) 6 ( 2 ) 4 To check, multiply ( + 2 ) ( 2 + 3 1 4 T + 2 ) The result should be + 5 6

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112 Example H ) Find the quotient: ( 8 + 27 ) ( 2 + 3 ) This time we will show the division all in one step. We need to add two placeholders in order to divide. ========================================================================== Worksheet : 5.7 Divid e Polynomials Divide a monomial by a monomial : 1 ) 20 ( 30) 2) 3 ) ( ) ( ) Divide a polynomial by a monomial: 4) ( 9 + 6 ) ( 3 ) 5 ) 6 ) Divide a polynomial by a polynomial (use long division): 7) ( + 7 + 12 ) ( + 3 ) 8) ( 2 35 ) ( + 5 ) 9) ( 4 – 17 15 ) ( 5 ) 10) ( + 11 + 16 ) ( + 8 ) 11) ( 3 + + 4 ) ( + 1 ) 12) ( 64 27 ) ( 4 3 ) 13) ( 9 ) ( + 3 ) ===================================================================== ( 8 + 27 ) ( 2 + 3 ) 4 6 + 9 2 + 3 8 + 0 + 0 + 27 ( 8 + 12 ) 12+ 0 ( 12 18 ) 18 + 27 ( 18 + 27 ) 0 4 ( 2 + 3 ) 6 ( 2 + 3 ) 9 ( 2 + 3 ) To check, multiply ( 2 + 3 ) ( 4 6 + 9 ) The result should be 8 + 27

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113 6.1 FactoringGreatest Common Factor Learning Objectives : In this section, you will: Find the Greatest Common Factor of a list of numbers. Find the Greatest Common Factor of a list of variable terms. Factoring the Greatest Common Factor of a Polynomial: When we study fractions, we learn that the greatest common factor (GCF) of two numbers is the largest number that divides evenly into both numbers. For instance, 4 is the GCF of 16 and 20 because it is the largest number that divides evenly into both 16 and 20. The CGF of polynomials works the same way: 4x is the GCF of 16x and 20x2. Because it is the largest polynomial that divides evenly into both 16x and 20x2 When factoring a polynomial expression, our first step should be to check for a GCF. Look for the GCF of the coefficients, and then look for the GCF of the variables. Find the GCF of a list of numbers. Example A) Find the GCF for 60, 45. Example B) Find the GCF for 20, 13,15 13=13 GCF=1 (there are no primes common to all three numbers, so the CGF is 1 Find the GCF of a list of variable terms. The terms x6, x11 have x6 AS the GCF since the smaller power on the variable x in the factored forms is 6. Example C) Find the GCF for x3y3z5 and x2yz10. GCF = x2yz5 Example D) Find the GCF for 60x3y3z5 and 45x2yz10. GCF= 15x2yz5 Factoring by GCF The opposite of multiplying polynomials together is factoring polynomials. There are many benefits of a polynomial being factored. To factor a number is to write it as a product of two or more numbers. When we multiplied polynomials, we multiplied monomials by polynomials by distributing, solving problems such as 4 ( 2 3 + 8 ) = 8 13+ 32 . Example E) Write in factored form by GCF : 10y2+35y3 since GCF=5y 10y2+35y3 = 5y(2y) +5(7y2) = 5y(2y+7y2)

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114 Check : 5y(2y+7y2) Distributive property, multiply the factored form. = 5y(2y) +5(7y2) = 10y2+35y3 Example F) 30m5+20m4+15m3 GCF=5m3 30m5+20m4+15m3 = 5m3(6m2) + 5m3 (4m) + 5m3(3) = 5m3(6m2+ 4m+3) Example G) 8x3+24x2+16x GCF = 8x 8x3+36x2+16x = 8x(x2) 8x(4x) 8x (2)= 8x(x2+4x 2) ================================================================== Worksheet 6.1: FactoringGreatest Common Factor Factor the common factor out of each expression. 1) 9 + 8 b2 3) 45 x2 25 5) 56 35p 7) 7ab 35a2b 9) 3a2b + 6 a3b2 11) 5x2 5x3 15 x4 13) 20x4 30x + 30 15) 28 m4 + 40 m3 + 8 2) x 5 4) 1 + 2 n2 6) 50x 80y 8) 27 x2y5 72x3y2 10) 8 x3y2 + 4 x3 12) 32n9 + 32n6 + 40 n5 14) 21 p6 + 30 p2 + 27 16) 10x4 + 20x2 + 12x

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115 6.2 FactoringGrouping Learning Objectives : In this section, you will: Factor polynomials with four terms using grouping. Factoring by Groping When a polynomial has 4 terms, the GCF might be used to factor by grouping. ( 2a+3) (5b+2) Distribute (2a+3) into second parenthesis. 5b(2a+3) +2 (2a+3) Distribute each monomial. 10ab+15b+4a+6 Our solution The solution has four terms in it. We arrived at the solution by looking at the two parts, 5b (2a + 3) and 2(2a + 3). When we are factoring by grouping, we will always divide the problem into two parts, the first two terms and the last two terms. Then we can factor the GCF out of both the left and right sides. When we do this our hope is what is left in the parenthesis will match on both the left and right. If they match, we can pull this matching GCF out front, putting the rest in parenthesis and we will be factored. The next example is the same problem worked backwards, factoring instead of multiplying. Example A) Factor 10ab+15b+4a+6 10ab+15b+4a+6 Split problem into two groups. 10ab+15b+4a+6 GCF on left is 5b, on the right is 2. 5b(2a+3) +2 (2a+3) (2a+3) is the same! Factor out this GCF. ( 2a+3) (5b+2) Our Solution (factored form) The key for grouping to work is after the GCF is factored out of the left and right, the two binomials must match exactly. If there is any difference between the two, we either have to do some adjusting or it can’t be factored using the grouping method. E xample B) Factor 5xy8x 10y+16 5xy8x 10y+16 Split problem into two groups. 5xy8x 10y+16 GCF on left is x, on right we need a negative, so we use 2 x(5y 8) 2(5y 8) ( 5y 8) is the same! Factor out this GCF. (5y 8) ( x 2) Our Solution Example C) Factor 6 15 + 2 5 6 15 + 2 5 Split problem into two groups. GCF on left is 3 x2 on right, no GCF, use 1. 3 ( 2 5 ) + 1 ( 2 5 ) ( 2x 5) is the same and facotor out this GCF (2x 5)(3x2+1) Our Solution

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116 Worksheet 6.2: Factoring Polynomial by Grouping Factor each completely. 1) 40r3 8r2 25r + 5 2) 35x3 10x2 56x + 16 3) 3n3 2n2 9n + 6 4) 14 v3 + 10 v2 7 v 5 5) 15b3 + 21b2 35b – 49 6) 6x3 48x2 + 5x – 40 7) 3 x3 + 15x2 + 2 x + 10 8) 28p3 + 21 p2 + 20 p + 15 9) 35x3 28x2 20x + 16 10) 7 n3 + 21 n2 5 n 15

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117 6.3 FactoringTrinomials where a=1 Learning Objectives : In this section, you will: Factor trinomials with coefficient a=1 Factor trinomials after factoring out the GCF. Factoring with three terms, or trinomials, is the most important type of factoring to be able to master. As factoring is multiplication backwards, we will start with a multiplication problem a nd look at how we can reverse the process. Distribute (x +6) through second parenthesis. Distribute each monomial through parenthesis x2 Combine like terms. x 2 Our Solution The trick to making these problems work is how we split the middle term. Why did we pick +6x and not how do we know what is the one combinati on that works? To find the correct way to split the middle term we will use what is called the ac method. In the next lesson we will discuss why it is called the ac method. The way the ac method works is we find a pair of numbers that multiply to a certain number and add to another number. Here we will try to multiply to get the last term and add to get the coefficient of the middle term. In the previous 1 example that would mean we Example A) Factor completely + 9 + 18 + 9 + 18 Want to multiply to 18, add to 9. + 6 + 3 + 18 6 and 3, split the middle term. ( + 6 ) + 3 ( + 6 ) Factoring by grouping ( 3 ) ( 1 ) Example B) Factor completely 8 20 8 20 Want to multiply to 20, add to 8 10 + 2 20 10 and 2, split the middle term. ( 9 10) + 2 ( 10 ) Factor by Grouping ( 10 ) ( + 2 ) Factored form Example C) Factor completely + 10 + 21 + 10 + 21 + 7 + 3 + 21 ( + 7 ) + 3 ( + 7 ) ( + 3 ) ( + 7 ) Example D) Factor completely m2 2 m2 2 – W Our Solution ==================================================================

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118 Worksheet 6.3: FactoringTrinomials where a=1 Factor each com pletely. (Remember to pull out the GCF first.) 1) p2 +17p + 72 2) x2 3) m2 9m + 8 4) x2 5) x2 6) x2 + 13x+ 40 7) a b2 + 12 a b + 32 a 8) 2 b2 34b+ 140 9) w2 7w t – 8t2 10) 4a2b – 28a b + 40b

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119 6.4 FactoringTrinomials where Learning Objectives : In this section, you will: Factor trinomials after factoring out the GCF. When factoring trinomials, we used the ac method to split the middle term and then factor by grouping. The ac method gets it’s name from the general trinomial equa tion, ax2 + bx + c, where a, b, and c are the numbers in front of x2, x and the constant at the end respectively. The ac method is named ac because we multiply a c to find out what we want to multiply to. In the previous lesson we always multiplied to ju st c because there was no number in front of x2. This meant the number was 1 and we were multiplying to 1c or just c. Now we will have a number in front of x2 so we will be looking for numbers that multiply to ac and add to b. Other than this, the process will be the same. Example A) Factor the trinomial 2x2 – x 6 by the grouping (“ac”) method. This is polynomial of the form ax2+bx+c. Determine the value of a, b and c. a=2, b= 1 and c= 6 Step1 , find “ac”: (2) ( 6) = 12 Step2 , find two integers whose product is “ac” and whose sum is “b.” The two integers are 4 and 3. Step 3 , Rewrite the middle term bx as the sum of two terms whose coefficients are integers found in the previous step2. Rewrite 2x2x 6 as 2x24x+3x 6 Step 4 , Factor by grouping 2x24x+3x 6 2x(x 2) +3(x 2) (x 2) (2x+3) Example B) Factor the trinomial 5x2 + 7x 6 5x2 + 7x 6 a=5, b=7 and c= 6. Determine ac= 30. We need to find two numbers with a product of 30 and a sum of 7. The numbers are 3 and 10, split the middle term. 5x23x+10x 6 x(5x 3) + 2(5x 3) (5x 3) (x+2) Example C) Factor the trinomial 3x2 + 11x + 6 3x2 + 11x + 6 Multiply to ac or (3)(6) =18, add to 11 3x2 + 9x + 2x + 6 The numbers are 9 and 2, split the middle term. 3x(x+3) +2(x+3) Factoring by grouping (x+3) (3x+2) Our solution When a = 1, or no coefficient in front of x2, we were able to use a shortcut, using the numbers that split the middle term in the factors. The previous example illustrates an important point, the problem.

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120 To factor using a GCF , take the greatest common factor (GCF), for the numerical coefficient. When choosing the GCF for the variables, if all terms have a common variable, take the ones with the lowest exponent. Example E) Factor the trinomial 9x4 3x3 + 12x2 9x4 3x3 + 12x2 GCF: Coefficients = 3 Variables (x) = x2 GCF = 3x2 Next, you just divide each monomial by the GCF! 3x2( 3x2 1 x 4) a=3, b= 1 and c= 4 use ac method ac= (3) ( 4) = 12 3x2(3x2 x 4) 3x2(3x2 +3x 4x 4) Factor by grouping 3x2{3x(x+1) 4(x +1)} 3x2(x+1) (3x +1) ================================================================== Worksheet 6.4 Factoring Trinomials where Factor Completely: 1)7x2 48x + 36 2) 7 b2 + 15 b + 2 3) 5a2 13a – 28 4) 2x2 5x + 2 5) 2x2 + 19x + 35 6) 2 b2 b 3 7) 5k2 + 13k + 6 8) 3x2 17x + 20 9) 3x2 + 17xy + 10 y2 10) 5x2 + 28 xy 49 y2 11) 6x2 39x – 21 12) 21k2 87k – 90 13) 14x2 60 x + 16 14) 6x2 + 29x + 20 15) 4k2 17k + 4 16) 4x2 + 9xy + 2 y2 17) 4m2 9 mn 9 n2 18) 4x2 + 13xy + 3 y2 19) 12x2 + 62xy + 70 y2 20) 24x2 52 xy + 8 y

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121 6.5 Factoring Special Products Learning Objectives : In this section, you will: Identify and factor special products including a difference of squares, perfect squares, and sum and difference of cubes. Factor trinomials after factoring out the GCF. When we learned how to multiply binomials, we talked about two special products: the Sum and Difference Formula and the Square of a Binomial Formula . Now, we will learn how to recognize and factor these special products. Factoring the Difference of Two squares: We use the Sum and Difference Formula to factor the difference of two squares. The difference of two squares is a quadratic polynomial in this form: a2b2. Both terms in the polynomial are perfect squares. In a case like this, the polynomial factors into the sum and difference of the square root of each term . = ( + ) ( ) Example A) Factor Subtracting two squares, the square roots are x and y (x+ y) (x y) Our solution Example B ) Factor 16 16 Subtracting two squares, the square roots are x and 4 (x+4) (x 4) Our solution Example C) Factor 9 25 9 25 Subtracting two squares, the square roots are 3a and 5b (3a+5b) (3a 5b) Our solution Example D) Factor + 36 + 36 No bx term, we use 0x. + 0 + 36 Multiply to 36, add to 0: No combinations that multiply to 36 add to 0 Prime, can’t factor. Example E) Factor Difference of squares with roots x2 and y2 ( + ) ( ) The first factor is prime, the second is x difference of y. ( + ) ( + ) ( ) Our solution Perfect Square: A Perfect square trinomial has the form a2 + 2ab + b2 = (a + b)2 or a2 2ab + b2 = (a b)2 Example A) Factor 6 + 9 6 + 9 Multiply to 9, add to 6 3 3 + 9 3 and 3, split the middle term. ( 3 ) 3 ( 3 ) Factoring by grouping ( 3 ) ( 3 ) ( 3 )

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122 Example B) Let’s use the formula. Factor 6 + 9 6 + 9 C heck the first term and the last term are perfect squares. Check the middle term is twice the product of the square roots of the first and the last terms. This is true also since we can rewrite them. 2 3 + 3 This means we can factor 6 + 9 as ( 3 ) Example C) Factor + 8 + 16 + 8 + 16 check the first term and the last term are perfect squares. Check the middle term is twice the product of the square roots of the first and the last terms. This is true also since we can rewrite them. + 2 . 4 + 4 This means we can factor + 2 4 + 16 as ( + 4 ) Factor a Sum/difference of cubes. Another factoring shortcut has cubes. With cubes we can either do a sum or a difference of cubes. Both sum and difference of cubes have very similar factoring formulas. Sum of Cubes : a3 + b3 = (a + b) (a2 2) Difference of Cubes : a3 b3 b) (a2 + ab + b2) Example A) Factor M3 M3 We have cube roots m and 3. 2 + 3m +9) Use formula Example B) Factor x3 x3 x3 3 we have cube roots x and 2. ( x x2 + 2 x + 22) Example C) Factor x3 x3 we have cube roots x and 4. ( x x2 + 4 x + 16) Example D) Factor 125p3+8r3 125p3+8r3 We have cube roots 5p and 2r. (5p + 2r) (25p 2 10r +4r 2) Use formula ==================================================================

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123 Worksheet 6.5: Factoring Special Products Factor:1) r2 2) p2 + 4 3) 9k2– 25 4) 3x2 5) 16x2 – 36 6) 18a2 2 7) a2 8) x2 + 6x + 9 9) x 2 6x + 9 10) 25p2 11) 25a2 + 30ab +9b2 12) 4a2 2 13) 8x2 24xy + 18y2 14) 15) x3 64 16) 125a3 – 64 17) 64x3 + 27y3 18) a4 19) 16 – z4 20) m4 4

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124 6.6 Factoring Strategies Learning Objectives : In this section, you will: Identify and use the correct method to factor various polynomials. With so many different tools used to factor, it is easy to get lost as to which tool to use when. Here we will attempt to organize all the different factoring types we have seen. A large part of deciding how to solve a problem is based on how many terms ar e in the problem. For all problem types we will always try to factor out the GCF first. Factoring Strategy (GCF First) 2terms: sum or difference of squares or cubes Difference of Squares a2 – b2 Sum of Squares a2 – b2 = Prime Sum of Cubes a3 + b3 = (a + b) (a2 2) Difference of Cubes a3 32 + ab + b2) 3terms: ‘ ac ’ method, watch for perfect square. Perfect Square a2 + 2ab + b2 = (a + b)2 or a2 2ab + b2 = (a b)2 Multiply to ‘ ac’ and add to ‘ b’ 4terms: grouping method We will use the above strateg ies to factor each of the following examples. Here the emphasis will be on which strategy to use rather than the steps used in that method. Example A) Factor 4x2+ 56xy + 196y2 4x2+ 56xy + 196y2 GCF first, 4 4(x2 + 14xy + 49y2) Three terms, try ac method, multiply to 49, add to 14; 7 and 7, perfect square! 4(x + 7y)2 Our Solution Example B) Factor 5x2 2 5x2 2 GCF first, 5x Four terms, try grouping. GCF is (x + 3), factor out (x+3) Our Solution Example C) Factor 100x2 100x2 GCF first, 100 Two terms, difference of squares Our Solution Example D) Factor 108x3 y2 2 y2 + 3xy2 108x3 y2 2 y2 + 3xy2 GCF first, 3xy2 3xy2 (36x2 3xy2 (36x2 3xy2 Factor by grouping 3xy2 Our Solution ==================================================================

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125 Worksheet 6.6: Factoring Strategies Factor : 2) 5u2 2 3 + 128y3 4) 5n3 +7n2 5) 54u3 7) x2 2 8) 9x2 25y2 9) m2 2 10) 36b22 + 24xc 11) 128 + 54x3 12) 2x3 + 6x22x 13) n3 + 7n2 + 10n 14) 27x3 – 64 15) 5x2 + 2x 16) 3k3 2 + 60k 18) 16x2 8xy + y2 19) 27m2 2 20) 9x3 + 21x2 2x 21) 2m2 2

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126 6.7 S ol ve by Factoring Learning Objectives : In this section, you will: Solve quadratic equation by factoring and using the zeroproduct rule. Zero Product Rule: The zero product rule tells us that if two factors are multiplied together and the answer is zero, then one of the factors must be zero. If ab = 0 then either a = 0 or b = 0 Solving Polynomial Equations by factoring and using the zeroproduct rule We have learned how to factor quadratic polynomials that are helpful in solving polynomial equations like ax2+bx+c=0 . Remember that to solve polynomials in expanded form , we use the following steps: Step 1: Rewrite the equation in standard form such that: Polynomial expression = 0. Step 2: Factor the polynomial completely. Step 3: Use the Zero Product Property to set each factor equal to zero. Step 4: Solve each equation from step 3. Example A) Solve 4x2 Rewrite: No need to rewrite because it is already in the correct form. Factor: 4x2 4x2 Factor by grouping ( One factor must be zero Set each factor equal to zero = 3 4 , = 1 Example B) Solve x2 x2 Rewrite: Set equal to zero by moving terms to the left x2 Factor using the ac method, multiply to 15, add to – 8 = 0 x Set each factor equal to zero x = 5 or x = 3 Our Solution Example C) Solve 4x2 = 8x 4x2 = 8x Set equal to zero by moving the terms to left 4x2 Factor out the GCF of 4x One factor must be zero Set each factor equal to zero x = 0 or 2 Our Solution

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127 2 Example D) Solve 2x3 2 + 24x =0 2x3 2 + 24x =0 Factor out the GCF of 2x Factor with ac method, multiply to 12, add to – 7 The – 4 Set each factor equal to zero x =0 or x=3 or x=4 Our Solutions Example E) Solve x2 + 4x = 5 x2 + 4x – 5 = 0 (x + 5) (x – 1) = 0 x + 5 = 0 x – 1 = 0 x = 5 or x = 1 Example F) Solve x281=0 (x+9) (x 9) =0 x+9=0 or x 9=0 x=9 or x= 9 ================================================================== Worksheet 6.7: So l ve by Factoring Solve: 1) (k – 7) (k+2) = 0 10) 7x + 17x 2) ( x 1) ( x + 4) = 0 11) 7 r2 r 3) 6x2 150 = 0 12) x2 6x = 16 4) 2n2 + 10n 28 = 0 13) 3 v2 + 7 v = 40 5) 7x2 + 26x + 15 = 0 14) 35 x2 + 120 x 6) 5n2 9n 2 = 0 15) 4 k2 + 18k 23 = 6 k 7 7) x2 4x 8 = 8 16) 9 x2 46 + 7x = 7x + 8x2 + 3 8) x2 5x 1 = 5 17) 2m2 + 19 m + 40 = 2m 9) 49p2 + 371p 163 = 5 18) 40p2 + 183 p 168 = p + 5 p

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128 7.1 Reduce Rational Expressions Learning Objectives : In this section, you will: Reduce ration expressions by dividing out common factors. Rational Expression: Any expression that can be written as the quotient of two polynomials. Remember that for a fraction, the denominator cannot be zero, so any values we substitute in for our variables cannot make the denominator zero. If we are given a value, we just substitute that value in for our variable to simplify the expression. To reduce a fraction, we know we remove what they have in common. This is the same procedure done for these new rational expressions since they are really the same idea as every fraction you have ever reduced. The key is to remember that to do this, the pr oblem must be factored first. Point: To reduce a ration expression, we must first factor and then remove what the numerator and denominator have in common Example A) Evaluate when = 1 ( ) ( ) ( ) Substitute 1 for a = Use order of operations to simplify Example B) Simplify ( ) ( ) Factor the numbers ( ) ( ) = Cross out the common factors Example C) Determine any excluded values 12 24 = 12 ( 2 ) Factor the denominator 12 ( 2 ) = 0 Set the denominator equal to zero to find excluded values 12 = 0 2 = 0 Set each factor equal to zero = 0 = 2 Our solution

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129 Example D) Simplify 3 + 2 = ( 2 ) ( 1 ) Factor the numerator + 5 14 = ( 2 ) ( + 7 ) Factor the denominator =( ) ( ) ( ) ( ) Rewrite the fraction in factored form ( ) ( ) ( ) ( ) = Cross out the common factor ================================================================== Worksheet 7.1: Reduce Rational Expressions Evaluate: 1) when = 5 2) when = 2 3) when = 3 State the excluded values for each: 4) 5) 6) Simplify the expression: 7) 8) 9) 10) 11) 12) 13) 14) 15) 16) 17) 18) 19) 20) 21)

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130 7.2 Multiply and Divide Rational Expressions Learning Objectives : In this section, you will: Multiply and divide rational expressions. The process of multiplying or dividing rational expressions is the same process we have used for fractions in the past. The only thing we need to do is factor the problem first before we actually multiply. Point: Remember all division is just multiplying by the reciprocal. Change the problem to multiplication first before factoring and simplifying. = Point: We can cancel any numerator factor with any denominator factor. This is called cross cancelling. Example A) Multiply: Cross cancel any numbers Cross cancel any variables Multiply across to get the final solution Example B) Divide Rewrite the problem as multiplication Cross cancel any numbers Cross cancel variables = Multiply across and simplify any remaining numbers

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131 Example C) Multiply ( ) ( ) ( ) ( ) Factor each numerator and denominator ( ) ( ) ( ) ( ) Cross cancel ( ) ( ) Multiply across to get the solution Example D) Divide ( } Rewrite as multiplication ( ) ( ) ( ) ( ) ( ) ( ) Factor each numerator and denominator ( ) ( ) ( ) ( ) ( ) ( ) Cross cancel ( ) ( ) Multiply across to get the solution ==================================================================

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132 Worksheet 7.2: Multiply and Divide Rational Expressions Multiply or Divide: 1) 2) 3) 4) 5) 6) 7) 8) 9) 10) ( ) 11) 12) 13) 14) ( ) 15) 16) 17) 18) 19) 20) ( 2 a 20a + 32 ) 21) ( 6 )

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133 7.3 Least Common Denominators Learning Objectives : In this section, you will: Find the least common denominator for rational functions Least Common Denominator : The smallest denominator that all other denominators can divide into. This is the same idea as a least common multiple, just in the denominator of a term. Since rational expressions are just fractions, we will find the least common denominator(LCD) very similarly to what we have always done. Often, a tree diagram can be helpful for numerical values. Step 1: Fact or the denominators completely Step 2: Looking at each denominator separately, count up how many of a given factor it has Step 3: Compare the factor count to all other factored denominators, whoever has the most of that factor, that is how many copies to take. We do not just take all the copies added up Step 4: Repeat this process until you have went through all different terms Whatever you have at the end is the LCD. Point : We want the most a term ap pears in any of the denominators. If a term appears the same amount in all denominators, that is the most so we will take that amount. Working with variables becomes easier when you see that x+1 has no x’s in it. Remember x and x+1 might as well be x and y. If x=2, then x+1=3, which have nothing in common. Anytime a variable has a number added or subtracted from it, it is an entirely different variable. Example A) Find the LCD: 15 and 12 3 5 and 2 3 Factor each denominator fully 3 5 2 Notice both numbers require one 3, the first number requires a 5 and the second number requires two 2s, so this gives us the numerical LCD Notice both terms require two x’s, so we take two. The first term needs one y while the other needs three, so we take three. 60 Solution

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134 Example B) Find the LCD: 5 + 5 and 25x 5 ( + 1 ) and 5 Factor each denominator fully 5 Notice the first term requires one 5, while the other requires two 5, so we take the greater amount ( + 1 ) Notice the first term requires an x+1, while the second requires an x, so we take each 25 ( + 1 ) Solution Example C) Rewrite both fractions over their LCD , ( + 1 ) ( 5 ) Find the LCD , Multiply each fraction by its missing factors ( ) ( ) , ( ) ( ) Multiply out the numerators Example D) ) Rewrite both fractions over their LCD , 7 x + 12 = ( 3 ) ( 4 ) Factor the first denominator 5 x + 6 = ( 2 ) ( 3 ) Factor the second denominator ( 3 ) ( 4 ) ( 2 ) Find the LCD ( ) ( ) , ( ) ( ) Multiply each fraction by its missing factors ( ) ( ) ( ) , ( ) ( ) ( ) Multiply out the numerators ==================================================================

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135 Worksheet 7.3: Least Common Denominators Find the LCD: 1 ) 3 x2y4 and 6 2) and + 5 3) 25 and 5 4) + 3 + 2, + 5 + 6 5) 4 8 , 2 , 4 6) + 2 + 1 , + 3 + 2 Write the fractions over their common denominator: 7) , 8) , 9) , 10) , , 11) , 12) , 13) , 14) , 15) , 16) , 17) , 18) , 19) , , 20) , 21) ,

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136 7.4 Adding and Subtraction Rational Expressions Learning Objectives : In this section, you will: Adding and subtracting rational expressions by having a least common denominator Adding or subtracting rational expressions is the same process as when you first learned to add and subt ractions in primary school. We need a common denominator, then we add or subtract the numerators and keep the denominator what it is. Remember we change the numerator when combining fractions with addition and subtraction, not the denominator. If you need practice finding the least common denominator(LCD), please review section 7.3. Point : We simplify fractions after we have done the addition or subtraction. If we simplify after finding the LCD but before combining the fractions, we will be back to where we started and cannot combine them. A few additional things to keep in mind. Once we combine our fractions through addition or subtraction, we do not multiply out the denominator. The reason is that if we can simply the final solution, we will need it factored to do that simplification. We do simplify, ie. distribute and collect like terms for the numerator. Once done, we factor the numerator if it can simplify with a factor in the denominator. If not, we do not need to factor the numerator. Example A) Add + LCD: 21 Find the LCD for every denom inator + Multiply each fraction by the factor needed to put them over the LCD + Combine terms after multiplying to get LCD Add the numerators together while writing the fraction over the LCD.

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137 Example B) Subtract 2 + 4 = 2 ( + 2 ) 4 = ( + 2 ) ( 2 ) Factor the denominators LCD: 2(x+2)(x 2) Find the LCD for every denominator ( ) ( ) ( ) Multiply each fraction by the factor needed to put them over the LCD ( ) ( ) ( ) ( ) ( ) Combine terms after multiplying to get LCD ( ) ( ) Add the numerators together ( ) ( ) Simplify Example C) Add + 4 + 3 = ( 3 ) ( 1 ) 2 3 = ( 3 ) ( + 1 ) Factor each denominator LCD: ( 3 ) ( 1 ) ( + 1 ) Find the LCD for every term ( ) ( ) + ( ) ( ) Multiply each fraction by the factor needed to put them over the LCD ( ) ( ) ( ) + ( ) ( ) ( ) Multiply out the numerators ( ) ( ) ( ) Add the numerators together Example D) Add and subtract + LCD: 2 ( ) Find the LCD ( ) ( ) + ( ) ( ) Multiply each fraction by the factor needed to put them over the LCD ( ) + ( ) ( ) Multiply out the numerators ( ) Add the numerators together ==================================================================

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138 Worksheet 7.4: Adding and subtracting rational expressions Evaluate: 1) + 2) 3) + 4) + 5) 6) + 7) 8) + 9) + 10) 4 + 11) 12) + 13) 14) + 15) + 16) + 17) + 18) + 19) 20) 21) 2 ( ) ( )

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139 7.5 Complex Fractions Learning Objectives : In this section, you will: Simplify complex fractions by multiplying each term by the least common denominator. Definition : A complex fraction is a fraction that has fractions in the numerator and/or denominator. As with all rational expressions, we need the problem to be factored before we work with it. Once i t is factored, our goal is to multiply every term by the least common denominator. Method 1. Find the LCD of every term. 2. Multiply every term by the LCD. 3. Reduce the fraction. 4. Simplify(distribute and collect like terms). This works because if we multiply every term by the LCD, we are multiplying by a term over itself, which means we are multiplying by 1. This method does not work for adding or subtracting fractions. Point : When this method is used correctly, every denominator will cancel away with the numerator. This removes all the denominators and gives us a problem we know now to solve. Example A) Simplify : 35 Fin d the LCD Multiply every term by the LCD ( ) ( ) = Simplify Example B) Simplify LCD: x+4 Find the LCD for every denominator ( ) ( ) ( ) ( ) ( ) Multiply every term by the LCD ( ) ( ) ( ) ( ) ( ) Reduce fractions = Simplify

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140 Example C) Simplify LCD: Find the LCD for every denominator ( ) ( ) Multiply each fraction by the LCD ( ) ( ) Reduce fractions Multiply ( ) ( ) ( ) Factor to find like terms to simplify ( ) ( ) ( ) = Reduce the fraction Example D) Simplify LCD: ( + 4 ) ( 3 ) Find the LCD for every denominator ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) Multiply every term by the LCD ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) Reduce fractions = Multiply and simplify ==================================================================

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141 Worksheet 7.5: Complex Fractions Solve: 1) 2) 3) 4) 5) 6) 7) 8) 9) 10) 11) 12) 13) 14) ( ) 15) 16) 17) 18) 19) 20) 21)

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142 7.7 Solving Rational Equations Learning Objectives : In this section, you will: Solve rational equations by multiplying by the LCD The toughest part of solving a rational equation is having fractions. If we can remove the fractions, we would have the sort of problems we have solved previously. Our technique will remove the denominator in order to give an easier problem to solve. If the problem is a simple fraction=fraction, we can cross multiply: = ad = bc For more complicated problems we will follow this method 1. Find the LCD of every term. 2. Multiply every term by the LCD. 3. Solve the resulting problem. 4. Check if any of the solutions make a denominator zero. If so, we must remove that solution. Remember to check the solution here. Since we are not solving the original problem but instead we are solving a similar problem, its possible to get solutions that do not work because they would make us divide by zero. These types of solutions are called e xtraneous solutions. We never loose solutions, but sometimes gain ones we cannot use. Point : When this method is used correctly, every denominator will cancel away with the numerator. This removes all the denominators and gives us a problem we know now t o solve. This method works since we are allowed to multiply both sides of an equation by the same term. This does not work in adding and subtraction rationals. Example A) Add = 3 ( 7 ) = 5 ( 2 ) Cross multiply 21 = 10 Simplify = Solve and check

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143 Example B) Subtract + = LCD: 42 Find the LCD for every denominator + = Multiply each fraction by the LCD + = Cross cancel individual terms 28 + 30 = 9 Multiply = Solve and check Example C) Solve = LCD: 5 ( + 2 ) Factor each denominator ( ) ( ) = ( ) Multiply each fraction by the LCD ( ) ( ) = ( ) Cross cancel individual terms 15 5 ( + 2 ) = + 2 Multiply 10 10 = + 2 Simplify = Solve and check

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144 Example D) Add and subtract = LCD: ( + 1 ) ( + 3 ) Find the LCD for every denominator ( ) ( ) ( ) ( ) ( ) ( ) = ( ) ( ) Multiply every term by the LCD ( ) ( ) ( ) ( ) ( ) ( ) = ( ) ( ) Cross cancel 2 ( + 1 ) ( 3 + 5 ) = 5 ( + 3 ) Multiply 3 = 5 + 15 Simplify = 3 , no solution Solve and Check Notice that 3 makes two of the denominators zero, so we cannot include it. That mean s there is no solution. ==================================================================

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145 Worksheet 7.7: Solving Rational Equations Solve: 1) = 2) = 3) = 4) = 5) = 6) = 7) + = 8) + = 9) = 1 10) = = 6 11) = 1 + 12) = + 13) = 14) + = 15) = 16) = 17) + = 18) = 19) = 20) =

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146 8.1 Square Roots and 8.2 Higher Roots Learning Objectives : In this section, you will: Simplify radical expressions. Square Root ( ) : The square root of a number is a value that, when multiplied by itself, gives the original number. Example A) Find 9 It represents the square root of 9, which is 3 because 3 multiplied by itself 3 * 3 equals 9. Cube Root ( ): The cube root of a number is a value that, when multiplied by itself twice (cubed), gives the original number. Follow similar procedures for higher roots, such as 4th ro ot, etc. Example B) Find 8 It represents the cube root of 8, which is 2 because 2 multiplied by itself twice 2 * 2 * 2 equals 8. Simplify radical means remove perfect square factors: If the number under the radical sign (the radicand) has perfect square factors, you can simplify by taking the square root of those factors out of the radical. Example C) Simplify 16 Recognize that 16 is a perfect square, and its square root is 4. Result: 16 = 4 Or Prime factorization of 16 = 24, 16 = 2 2 2 2 = 2 2 = 4 Example D) Simplify 64 Recognize that 64 is a perfect cube, and its cube root is 4. Result: 64 = 4 Or Sometimes it is not easy to see the perfect cubes, in this case find the prime factorization of 64 = 26, 64 = 2 2 2 2 2 2 = 2 2 = 4 Example E) Simplify 50 Step 1: Recognize that 50 is not a perfect square, but it contains a perfect square factor, which is 25. Step 2: Rewrite 50 as 25*2. Step 3: Take the square root of 25, which is 5. Result: 50 = 25 2 = 5 2 Or Use prime factorization if you cannot find perfect square: 50 = 5 5 2 = 5 2 In prime factorization it is easy to see which identical two numbers’ product is a perfect square (that is the number that will move outside of the square root).

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147 Example F) Simplify 48 Step 1: Recognize that 48 is not a perfect cube, but it contains a perfect cube factor, which is 8. Step 2: Rewrite 48 as 8 * 6. Step 3: Take the cube root of 8, which is 2. Result: 48 = 2 6 Or Use prime factorization if you cannot find perfect square: 48 = 2 2 2 2 3 = = 2 2 3 =2 6 (in this case look for the product of three identical numbers perfect cube and that number will move outside of the cube root) Example G) Simplify Use same strategy with variables as with numbers: use perfect squares or prime factorization: = = Example H) Simplify 32 32 = 2 2 2 2 2 = 2 2 =============================================== ====================== Worksheet: 8.1 Square Roots and 8.2 Higher Roots 1) Evaluate: 64 2) Simplify: 75 3) Simplify: 4 72 4) Simplify: 54 5) Simplify: 8 600 6) Express 98 in simplest radical form. 7) Express 50 in simplest radical form. 8) Simplify : 125 9) Express 27 in simplest radical form. 10) Express 2 48 in simplest radical form.

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148 8.3 Adding Radicals Learning Objectives : In this section, you will: Add and Subtract radical expressions with and without variables Adding radicals as like terms involve s combining expressions with the same radicand (the number inside the radical symbol) and the same index (the root). The key is to combine the coefficients while preserving the radicand and the index. 1. Simplify your radical expressions. 2. If the radicands an d indices match, you can treat them as like terms and perform the addition or subtraction accordingly. (If the radicands or indexes do not match you cannot combine your radical expressions.) Example A) Add: 3 + 3 Explanation: When you add like terms, you add the coefficients (numbers outside the radicals) while keeping the radicand and the index the same. Solution: 2 3 Example B) Add: 2 5 + 4 5 index (2) the same. Soluti on: 6 5 Example C) Subtract: 4 2 2 Explanation: Subtract the coefficients (4 1) while keeping the radicand ( 2) and index (3) the same. Solution: 3 2 Example D) Simplify: 4 5 3 2 + 5 + 2 2 Combine like radicals. 4 5 + 1 5 = 5 5 and 3 2 + 2 2 = 1 2 Solution: 5 5 2 (Cannot combine not like radicals.) Example E) Add: 8 + 32 Explanation: Simplify each radical first: 8 = 2 2 and 32 = 4 2 . Then, add the like terms, 2 8 + 4 2 = 6 2 Solution: 6 2

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149 Example F) Simplify: + Explanation: Here, you cannot simplify further since the radicands ( and ab) are different. Solution: + Example G) Simplify: 2 1) Explanation: Explanation: Simplify each radical first: you cannot simplify the first, = 2) 2 3) Subtract the coefficients : 2 – 1 = 1 while keeping the radicand and index : 3 the same. 4) Solution: Example H) Simplify: 8 5 45 80 Simplify the radicals: 8 5 3 3 5 2 2 2 2 5 = 8 5 3 5 4 5 Combine like radicals. Solution: 5 ===================================================================== Worksheet: 8.3 Adding Radicals Simplify and add/subtract: 1) 2 3 + 3 3 2) 3 + 2 3) 4 3 3 4) 20 + 3 5 5) 4 16 54 6) 4 3 5 7 + 5 3 + 2 7 7) 3 18 + 3 50 8) 2 4 9) 4 50 + 5 27 3 2 2 108 10) 5 + 3 5 8 5 + 2 5 11) 81 3 32 + 24

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150 8.4 Multiplying Radicals Learning Objectives : In this section, you will: Multiply radical expressions with and without variables "Product Rule" for radicals: when you multiply two radicals with the same index (root), you can simplify the expression by multiplying the contents (the numbers or variables) inside the radicals together under a single radical. Here's the rule in mathematical notation: a b = a b = ab In words, this rule tells us that the product of two radicals ( a and b ) is equal to the square root of the product of their contents ab . If the radicals have coefficients, multiply coefficients separately and multiply the radicals. a b = a b = ab Here's a breakdown of the steps for multiplying radicals: Example A) Multiply Radicals with Numbers: 5 10 1. Multiply the numbers inside the radicals: 5 10 = 50 2. Simplify the result: 2 5 5 = 5 2 (you cannot simplify further) Example B) Multiply Radicals with Coefficients: 2 3 3 5 1. If there are any coefficients or constants outside the radicals, you should multiply them together. 2 3 3 5 , you can multiply 2 and 3 to get 6. 2. Then, you multiply the contents inside the radicals together. In our example, 3 5 becomes 3 5 = 15 . 3. If further simpl ification is possible (e.g., simplifying the radical), you should do that. In this case, 15 cannot be simplified further. 4. So, 2 3 3 5 simplifies to 6 15 . Example C) Multiply Radicals with Variables: a 1. Multiply the variables inside the radicals: a = Take the square root of the result: 2. Simplify: = b b b = b Example D) Multiply Radicals with Numbers and Variables: 3x 6y 1. Multiply the numbers and variables inside the radicals: 3x 6y = 18xy 2. Take the square root of the result: 18xy 3. Simplify the radical: 2 3 3xy = 3 2xy

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151 Example E) Multiply Radicals with Variables with Exponents: 1. Multiply the variables with exponents inside the radicals: = ( ) = = 2. Take the square root of the result, simplify: = Example F) Distribute Radicals (multiply 1 term by 2 terms): 2 2 5 10 + 14 1. Distribute to both terms: 2 2 5 10 + 14 = 10 20 + 2 28 2. Simplify both radicals: 10 20 + 2 28 = 10 4 5 + 2 4 7 = 20 5 + 4 7 Example G) Distribute Radicals (multiply 2 terms by 2 terms): 2 2 + 4 6 + 5 1. Distribute using FOIL: 2 2 + 4 6 + 5 = 2 12 + 10 2 + 4 6 + 20 2. Simplify: 2 12 + 10 2 + 4 6 + 20 = 2 4 3 + 10 2 + 4 6 + 20 = 4 3 + 10 2 + 4 6 + 20 Example H) Distribute: 6 + 7 7 1. Distribute using FOIL: 2 2 + 4 6 + 5 = 2 12 + 10 2 + 4 6 + 20 2. Simplify: 2 12 + 10 2 + 4 6 + 20 = 2 4 3 + 10 2 + 4 6 + 20 = 4 3 + 10 2 + 4 6 + 20 ===================================================================== Worksheet: 8.4 Multiplying Radicals Multiply: 1) 2 2 2) 2 3 5 6 3) 3 5 4) 3 5 6 5) 3 8 7 10 6) 5 3 5 3 7) 40 4 8) 11 10 11 9) 2 + 5 10 + 5 10) 6 5 6 + 5 11) + x 12) 5 + 14 7 5 + 7 13) 5 7

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152 8.5 Dividing Radicals, Rationalizing Denominator Learning Objectives : In this section, you will: D ivide radical expressions with and without variables D ivide by monomial (one term) D ivide by binomial (two terms) R ationalize denominator Dividing Radical Expressions (Quotient Rule): "Quotient Rule" for radicals: when you divide two radicals with the same index (root), you can simplify the expression by dividing/simplifying the contents (the numbers or variables) inside the radicals togethe r under a single radical. Here's the rule in mathematical notation: = In words, this rule tells us that the quotient of two radicals : and is equal to the square root of the quotient of their contents . If the radicals have coefficients: divide/simplify coefficients separately and divide/simplify the radicals. Example A ) Divide Numbers: 1) Divide the numbers inside the radicals: = = 4 2) Take the square root of the result: 4 = 2 Example B) Divide Variables with Exponents: 1) Divide the variables inside the radicals: = 2) Simplify (subtract exponents): = = 3) Simplify the radical: = Example C) Divide Numbers and Variables: 1) Simplify coefficients: = 2) Divide the numbers and variables inside the radicals: = 4 3) Take the square root of the result (do not forget the coefficients) and simplify: = =

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153 Example D) Simplify the radical expression (multiple terms in numerator): 1) Simplify numerator: = 2) Factor numerator = ( ) 3) Simplify by dividing numerator and denominator by GCF: ( ) = ( 2 3 ) Rationalizing Denominator Dividing Radical Expressions by Monomial Rationalizing the denominator is a mathematical process used to simplify or eliminate radicals (square roots, cube roots, etc.) from the denominator of a fraction. To rationalize the denominator, you typically multiply both the numerator and denominator of a fraction by a carefully chosen expression that eliminates the radical from the denominator. Example E) Rationalize : 1) Rationalized Form: multiply both the numerator and denominator by 2 (the conjugate of the denominator 2 ) to rationalize the denominator. = = = 2) Simplify: = The simplified form is . (Realize: there is not any radical in the denominator.) Example F) Divide Numbers with Different Radicals: 1) Divide the numbers inside the radicals: = 2) Take the square root of the result: = = = 3) Rationalize the denominator by multiplying both the numerator and denominator by 3 : = = 4) Simplify: Rationalizing Denominator Dividing Radical Expressions by Binomial (Use Conjugate) The main goal is to eliminate the radical from the denominator. This often involves using the difference of squares formula, which states that (a + b)(a b) = a b. To rationalize the radical, you multiply both the numerator and the denominator of the expressi on by the conjugate of the denominator. For example, if you have the expression a

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154 Example G ) Rationalize the denominator (binomial expression in denominator): 1) Multiply by the conjugate of the denominator: Multiply both the numerator and denominator by ( 2 + 2 ), which is the conjugate of ( 2 + 2 ): 2) Simplify the expression: 2 + 3 2 2 ( 2 + 2 ) 2 + 2 = 4 + 2 2 + 2 3 + 6 4 + 2 2 2 2 4 = 4 + 2 2 + 2 3 + 6 4 4 = 4 + 2 2 + 2 3 + 6 4 2 = 4 + 2 2 + 2 3 + 6 2 Realize: there is not any radical in the denominator. ===================================================================== Worksheet: 8.5 Dividing Radicals, Rationalizing Denominator Divid e and rationalize the denominator: 1) 2) 3) 4) 5) 6) 7) 8) 9) 10) 11) 12) 13) 14)

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155 9.1 Quadratics Solving with Radicals Learning Objectives : In this section, you will: Solve equations with radicals. Check for extraneous solutions. An equation that contains a variable expression in a radical is a radical equation. + = ; n is called Index or Root. Solving equations requires isolation of the variable. Equations that contain a variable inside of a radical require algebraic manipulation of the equation so that the variable “comes out” from underneath the radical(s). This can be accomplished by raising both sides of the equation to the “nth” power, where n is the “index” or “root” of the radical. When the index is a 2 (i.e. a square root), we call this method “squaring both sides.” Sometimes the equation may contain more than one radical expression, and it is possible that the met hod may need to be used more than once to solve it. When the index is an even number (n = 2, 4, etc.) this method can introduce extraneous solutions, so it is necessary to verify that any answers obtained actually work. P lug the answer(s) back into the ori ginal equation to see if the resulting values satisfy the equation. It is also good practice to check the solutions when there is an odd index to identify any algebra mistakes. General Solution Steps: Step 1 . Isolate the Radical(s) and identify the index (n). Step 2 . Raise both sides of the equation to the “nth” power. Step 3 . Use algebraic techniques (i.e., factoring, combining like terms,) to isolate the variable. Repeat Steps 1 and 2 if necessary. Step 4 . Check answers. Eliminate any extraneous soluti ons from the final answer. Example A) Solve 5 3 = 0 Step1: Isolate the Radical 5 = 3 Step 2: Square both Sides ( 5 )= 3 Step 3: Solve for x 5 = 9 = 4 = 4 Step 4: Check Answers 5 ( 4 ) 3 = 0 9 3 = 0 , then 3=3, so x= 4 is the solution.

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156 Example B) Solve : 7 + 2 = 4 7 + 2 = 4 Square both s ides 7 + 2 = 16 Solve for x. 7 = 14 Divide both sides by 7. = 2 Check answer. 7 ( 2 ) + 2 = 4 16 = 4 So, x=2 is the solution. Example C) Solve 1 = 4 1 = 4 Odd index, we don’t have to check answer. 1 = ( 4 ) Cube both sides, simplify exponents. 1 = 64 Solve for x. = 63 This is the solution. Example D) Solve for 2 + 4 = 0 2 + 4 = 0 Isolate the radicals so that they are on opposite sides of the equal sign. 2 = + 4 Squa re both Sides ( 2 ) = ( + 4 ) 2 = + 4 Solve for x. 6 = 0 Factor ( + 2 ) ( 3 ) = 0 = 2 , = 3 Check answers. ( 2 ) 2 ( 2 ) + 4 = 0 if x= 2 4 2 ( 2 ) + 4 = 0 4 2 = 0 So, x= 2 is a solution ( 3 ) 2 ( 3 ) + 4 = 0 if x=3 9 2 ( 3 ) + 4 = 0 7 7 = 0 So, x=3 is also solution

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157 Example E) Solve for 6 + 9 + 3 6 + 9 + 3 = 0 Isolate the radicals so that they are on opposite sides of the equal sign. 6 + 3 = + 9 Square both s ides . ( 6 + 3 )= ( + 9 ) Recall: ( )= 2 + The first term will need to be distributed using FOIL . : 6 + 3 6 + 3 = 6 + 3 6 + 3 6 + 9 6 + 6 6 + 9 = + 9 Radical still remains, repeat step1 and step 2 6 = 1 ( 6 ) = 1 Square both s ides again. 6 = 1 Solve for x. = 7 Check the answer, and this is the solution. Example F) Solve for x: + 4 + 1 = 5 + 4 + 1 = 5 Isolate radical by subtracting x from both sides. 4 + 1 = 5 Square both side s. ( 4 + 1 ) = ( 5 ) U se Foil method ( )= 2 + 4 + 1 = 25 10 + Re order terms . 14 + 24 = 0 Factor . ( 12 ) ( 2 ) = 0 Set each factor equal to zero. 12 = 0 2 = 0 Solve each equation. = 12, = 2 Check answers in original problem. 12 + 4 ( 12) + 1 = 5 if x=12 , 12 + 48 + 1 = 5 12 + 7 = 5 19=5 False, extraneous root so x=12 is not an answer. 2 + 4 ( 2 ) + 1 = 5 if x=2 2 + 8 + 1 = 5 2 + 3 = 5 5=5 True. x =2 is the solution.

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158 Worksheet 9.1: Quadratics Solving with Radicals 1) 3 = 5 2) 2 + 3 3 = 0 3) 6 5 = 0 4) 3 + = 6 + 13 5) 3 3 1 = 2 6) 4 + 5 + 4 = 2 7) 2 + 2 = 3 + 2 1 8) + 4 = + 10 9) 7 4 = 2 3 10) 2 2 = 2 11) 1 = 2 + 1

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159 9.2 Solving with Exponents using the Square Root Property Learning Objectives: In this section, you will: Solve quadratic equations of the form x2 = k using the Square Root Property Solve quadratic equations of the form a(x –h)2=k using the Square Root Property We have already solved some quadratic equations by factoring. Let’s review how we used factoring to solve t he quadratic equation x2 = 9. Example A) Solve x2 = 9 using factoring: x2 9 = 0 Set to 0. (x 3)(x+3) = 0 Factor. x 3 = 0 or x+3 = 0 Set each factor to zero. x = 3 or x = 3 Solve for x. You have two solutions: x = 3, 3 Factoring can be used if ‘k’ is a perfect square. In case of any ‘k’ (perfect square or not) we can use the Square Root Property. Use the Square Root Property : If x2 = k , then x is a positive or negative square root of k. = Given a quadratic equation with an x2 term but no ‘x’ term, use the square root property to solve it. 1. Isolate the x2 term on one side of the equal sign. 2. Take the square root of both sides of the equation, putting a sign before the expression on the side opposite the squared term. 3. Simplify the numbers on the side with the sign. Example B) Solve a Simple Quadratic Equation Using the Square Root Property: x2 = 9 Solution: x2 = 9 Take the square root of both sides. Remember to use a sign before the radical symbol. = Then simplify the radical. = So, the solutions are = Realize that we got the same solutions as with factorization. Example C) Solve a Simple Quadratic Equation Using the Square Root Property: x2 = 8 Solution: x2 = 8 Take the square root of both sides. Remember to use a sign before the radical symbol. = Then simplify the radical. = So, the solutions are = Note: Realize that this problem cannot be solved with factorization over rational numbers.

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160 Example D) Solve the Quadratic Equation Using the Square Root Property: ( x 4 )= 36 Solution: ( )= Take the square root of both sides. Remember to use a sign before the radical symbol. ( ) = Then simplify the radical. = Split and solve the first degree equations. = or = Solve. x = 6+4 or x = 6+4 So, the solutions are = Note: Realize that this problem could be also solved by distributing, setting to zero, and factoring, since the roots are real numbers. Example E) Solving Using the Square Root Property: 4 ( a + 6 )+ 20 = 720 Solution: 4 ( a + 6 )+ 20 = 720 Isolate the square term. ( + )= Subtract 20 from both sides. ( + )= ( ) = Divide both sides by 4 ( + )= Take the square root of both sides. Remember to use a sign before the radical symbol. ( + ) = Take the square root of both sides. Remember to use a sign before the radical symbol. + = Then simplify the radical. = = Solve the first degree equations (since you cannot combine like terms, you can just subtract 6 from both sides and you do not have to split your solutions). = + = Another way to write your solutions. Realize that your answers are irrational numbers, so this problem cannot be solved with factorization over rational numbers. ===================================================================== Worksheet: 9.2 Solving with Exponents using the Square Root Property Solve the equations using the Square Root Property: 1) ( 4 )= 36 2) ( 2 )= 75 3) 3 ( 2 )= 48 4) 2 ( + 5 )= 16 5) ( + 4 )= 9 6) 4 ( 7 ) 10 = 350 7) ( 2 + 4 )= 49 8) ( 2 + 5 )= 27 9) 3 ( 3 7 ) 6 = 21 10) 2 2 = 18

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161 9.3 Completing the Square Learning Objectives: In this section, you will: Solve quadratic equations of the form a x2 + bx +c = 0 by completing the square All equations cannot be factored, only that ones that have rational roots. Consider the following equation: x2 there are two irrational solutions to this equation, 1 + 2 2 and 1 2 2 . To find these two solutions we will use a method known as completing the square. When completing the square, we will change the quadratic into a perfect square which can easily be solved with the square root property learned in 9.3. Example A) Solve the quadratic equati on by completing the square: x2 2x 7 = 0 1. Start with the quadratic equation: x2 2x 7 = 0 2. Move the constant term (in this case, 7) to the other side of the equation by adding it to both sides: x2 2x = 7 3. To complete the square, we need to make the coefficient of the x2 term (in this case, 1) equal to 1. Since it's already 1, we move to the next step. 4. Focus on the coefficient of the x term (in this case, 2). Divide it by 2 and square the result: ( 2 / 2)2 = ( 1)2 = 1 5. Add this value to both sides of the equation: x22x + 1 = 7 + 1 6. Now, you have a perfect square trinomial on the left side: (x 1) 2 = 8 7. Take the square root of both sides to solve for x: x 1 = 2 2 8. Solve for x: x = 12 2 So, the solution to the equation x2 2x 7 = 0 is x = 12 2 . Steps Now we can solve a Quadratic Equation (a x2 + bx +c = 0) in 5 steps by completing the square: Step 1 Divide all terms by a (the coefficient of x2). Step 2 Move the number term ( c/a ) to the right side of the equation. Step 3 Complete the square on the left side of the equation and balance this by adding the same value to the right side of the equation. We now have something that looks like (x + p)2 = q, which can be solved this way: Step 4 Take the square root on both s ides of the equation. Do not forget +/ solutions. Step 5 Subtract the number that remains on the left side of the equation to find x.

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162 Example B) Solve : 5x2 – 20x – 25 = 0 Step 1 Divide all terms by 5: x2 – 4x – 5 = 0 Step 2 Move the number term to the right side of the equation: x2 – 4x = 5 Step 3 Complete the square on the left side of the equation and balance this by adding the same number to the right side of the equation: (b/2)2 = ( 4/2)2 = ( 2)2 = 4 x2 – 4x + 4 = 5 + 4 (x – 2)2 = 9 Step 4 Take t he square root on both sides of the equation: x – Step 5 Add 2 to both sides, so the solutions are: x = 3 + 2 = 5 or 1 (Note: since the solutions are rational numbers, the equation could be solved by factorization.) Example C) Solve by completing the square: 2 36 = 98 Step 1 Divide all terms by 2: x2 – 18x = 49 Step 2 The number term is already on the right side of the equation: x2 – 18x = 49 Step 3 Complete the square on the left side of the equation and balance this by adding t he s ame number to the right side of the equation: (b/2)2 = ( 18/2)2 = ( 9)2 = 81 x2 – 18x + 81 = 49 + 81 (x – 9)2 = 32 Step 4 Take the square root on both sides of the equation: x – 9 = 32 and simplify: x – 9 = 4 2 Step 5 Add 9 to both sides, so the solutions are: x = 9 4 2 (Note: since the solutions are irrational numbers, the equation could not be solved by factorization.) =====================================================================

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163 Worksheet: 9.3 Completing the Square 1) Find the last term to make the trinomial into a perfect square: + 10 + _____ When this is factored, it becomes: 2) Find the last term to make the trinomial into a perfect square: 8 + _ _ _ _ _ _ _ _ When this is factored, it becomes: 3) Solve by completing the square: + 12 = 13 4) Solve by completing the square: = 6 + 1 5) Solve by completing the square: 2 36 + 160 = 0 6) Solve by completing the square: 4 = 16 + 20 7) Solve by completing the square: = 8 + 15 8) Solve by completing the square: + 2x = 8 + 89

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164 9.4 Quadratics Quadratic Formula Learning Objectives : In this section, you will: Solve quadratic equations by using the quadratic formula. The quadratic formula is a fundamental mathematical expression used to find the solutions or roots of a quadratic equation. A quadratic equation is a seconddegree polynomial eq uation, typically written in the form: + + = In this equation: “a” “ x ” represents the variable we want to solve for. The goal is to find the values of “ x ” that satisfy the equation, making it equal to zero. The quadratic formula is expressed as: = In the formula, the symbol means that there are two possible solutions, one with the positive square root and one with the negative square root. We can use the quadratic formula to solve any quadratic, this is shown in the following examples . Example A) Solve + + = using the Quadratic Formula. = ( ) ( ) ( ) Substitute a=1, b=3, and c=2 = Evaluate exponent and multiplication. = Evaluate root. = Evaluate to get two answers. = = Simplify fractions. = 1 = 2 Our Solution. Example B) Use the Quadratic Formula to solve for x : + = = ( ) ( ) ( ) ( ) Substitute a=3, b= 7, and c=2 = Evaluate exponent and multiplication. = Evaluate root. = Evaluate to get two answers. = = Simplify fractions. = 2 = Our Solution.

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165 Example C) Solve = using the Quadratic Formula. = ( ) ( ) ( ) ( ) ( ) Substitute a=2, b= 5, and c= 3 = Evaluate exponent and multiplication. = Evaluate root. = Evaluate to get two answers. = = Simplify fractions. = 3 = Our Solution. Example D) Solve = using the Quadratic Formula. = ( ) ( ) ( ) ( ) ( ) Substitute a=3, b=0 (missing term), and c= 7. = Evaluate exponent and multiplication. = Simplify root. = Evaluate to get two answers. = = Simplify fractions. = = Our Solution. Example E) Solve = using the Quadratic Formula. 4 12 = 9 First set equation equal to zero 4 12 + 9 = 0 Add 9 to both sides. = ( ) ( ) ( ) ( ) ( ) S ubstitute a=4, b= 12 and c=9. = Evaluate exponent and multiplication. = Evaluate root. = Evaluate to get two answers. = Simplify fractions. = Our Solution.

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166 Example F) Solve + = using the Quadratic Formula. 3 + + 7 = 0 First set equation equal to zero = ( ) ( ) ( ) ( ) ( ) Substitute a=3, b=1, and c=7. = Evaluate exponent and multiplication. = Simplify root. = 83 doesn’t represent a real number. Since 83 doesn’t represent a real number, there is no real number solution. ===================================================================== Worksheet 9.4: Quadratics Quadratic Formula 1) 4 + 4 = 0 2) 3 + 6 + 3 = 0 3) 2 7 + 3 = 0 4) + 5 + 6 = 0 5) 3 + 2 = 1 6) 2 3 = 0 7) 2 4 = 0 8) 5 10 + 5 = 0 9) 2 + 3 + 1 = 0 10) 3 4 + 2 = 0

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167 Beginning Algebra Answer Key for the Worksheets 0.1 Integers 1) 100 2) 20 3) 5 4) > 5) > 6) > 7) < 8) 5 9) 14 10) 2 11) 30 12) 9 13) 16 14) 6 15) 38 16) 4 17) 61 18) $330 19) 28 20) 40 21) 15 22) 6 23) 4 24) 2 0.2 Fractions 1) 2) 3) 17 4) 2 5) 6) 7) 1 8) 9) 10) 8 11) 12) 35 13) 5 14) 15) 27 16) 17) 18) 32 19) 20) 21) 22) 23) 24) 25) 26) 0.3 Order of Operations 1) 34 2) 7 3) 33 4) 102 5) 8 6) 5 7) 8) 9) 26 10) 11) 12) 13) 14) 15) 16) 5 0.4 Properties of Algebra (Simplify, Evaluate, Translate Expressions) 1) 32 2) 5 3) 6 4) 34 5) 34 6) 12 /7 7) 9 8) 7 9 9) 10) 10 + 3 11) 24 + 27 12) 5 – 9 13) 10 – 20 14) 2 – 2 15) 10 + 1 16) 14 + 90 17) 60 – 7 18) 11 19) – 8 20) + 7 21) / 22) 23) 3 + 24) 3 ( + )

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168 25) 7 + 26) 27) 2 28) 2 29) 5 – 2 30) 2 + 3 = 5 31) + 32) ( + ) 1.1 Solving Linear Equations One Step Equations 1) 7 2) 11 3) 5 4) 4 5) 10 6) 6 7) 19 8) 6 9) 18 10) 6 11) 20 12) 7 13) 108 14) 5 15) 8 16) 4 1.2 Linear Equations Two Steps Equations 1) 4 2) 7 3) 14 4) 2 5) 10 6) 12 7) 0 8) 12 9) 10 10) 16 11) 4y + 11 = 5; y = 4 12) 8 + 5x = 25; x = 14 13) 5 + 4x = 25; x = 5 14) .75x + 2.35 = 10; x = 10.2 mi 15) 3x + 150 = 300; x = 50 guests 1.3 General Linear Equations Multi Steps Equations 1) b = 2, conditional equation 2) x = , conditional equation 3) m = 3, conditional equation 4) y = 0, conditional equation 5) m = 3, conditional equation 6) y = 5, conditional equation 7) p = 4, conditional equation 8) all real numbers, identity 9) all real numbers, identity 10) no solution, contradiction 1.4 Solving with Fraction s 1) n = 2) k = 3) n = 0 4) b = 2 5) r = 1 6) p = 7) v = 8) x = 9) x = 10) x = 11) x = 13 1.5 Formulas 1) c = b a 2) x = g + f 3) L = S 2B 4) x = 5) L = 6) h = 7) T = 8) r = 9) w = 10) h = 11) v = 12) k = qr + m

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169 1.8 Application: Number/Geometry Number Problems 1) x=6 2) x=5 3) x= 4 4) x=32 5) x= 13 6) x=62 7) x=16 8) Son = $20 & Mr. Brown = $200 9) Boys = 15 & girls = 30 10) 14ft and 16ft 11) $1644 Geometry Problems 1) The first angle 56, the second angle 56, & the third angle 68 2) The first angle 64, the second angle 64, & the third angle 52 3) The first angle 30, the second angle 120, & the third angle 30 4) The first angle 40, the second angle 80, & the third angle 60 5) W = 30, L = 45 6) W = 56, L = 96 7) W = 57, L = 83 8) W = 17, L = 31 9) W = 112, L = 192 1.9 Other Applications: Age, Sales Tax, Discount, and Commission Problems 1) Boy = 16 & brother = 6 2) Son = 10 & father = 40 3) (a) The sales tax is $20.50 & (b) the total cost is $270.50 (b) 4) 9% 5) 7.5% 6) $ 273 7) $394.20 8) 6% 9) 4% 10) $450 11) (a) $11.60, (b) $17.40 12) (a) $ 256.75 (b) $138.25 3.1 Solve and Graph Inequalities 1) ( 2) 3) ( 2] 4) ( 5) ( 6) ( 7) x < 2 8) x 1 9) x 5 10) ( 11) ( 12) 13) [ 14) 15) 16) ( 17) [ 2.1 Graphing: Points and Lines Plot Points 1) B 2) D 3) O 4) H 5) C 6) F 7) ( 3, 2) 8) (1, 6)

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170 9) (8, 0) 10) (7, 8) 11) ( 8, 0) 12) (5, 5) 19) IV 20) III 21) II 22) I Graphing 1) (0, 3); (4, 45); ( 2, 27) 2) (0, 3); (2, 0); ( 2, 6) 2.2 Slope 1) 3/5 2) 2/3 3) 3/5 4) 5/11 5) 1/16 6) 12/31 7) 1/16 8) x int: (2, 0) ; y int: (0, 6) 9) x int: ( 5, 0) ; y int: (0, 5) 10) x int: (20, 0) ; y int: (0, 5) 11) x int: ( 5 , 0) ; y int: (0, 3) 12) x int: (4, 0) ; y int: (0, 12) 13) x int: (0, 0) ; y int: (0, 0) 14) x int: (5, 0) ; y int: none 2.3 Slope Intercept Form 1) Slope: 2/3; yint: (0, 4) 2) Slope: 1; yint: (0, 5) 3) Slope: 3/5; yint: (0, 1) 4) Slope: 53; yint: (0, 6) 5) Slope: 4/5; y int: (0, 8/5) 6) Slope: 4; yint: (0, 9) 7) y = 2x + 5 8) y = x – 4 9) y = 3x – 1 10) y = 13 x + 1 11) 12) 2.4 Point Slope Form 1) y + 4 = ( x 3 ) 2) y 4 = ( x + 1 ) 3) y 4 = 2 ( x 2 ) 4) y = 3 5) x = 6 6) y 6 = ( x 3 ) 7) y 1 = ( x 3 ) 8) y 4 = ( x 1 ) 9) x = 5 10) y = 3 2.5 Parallel & Perpendicular Lines 1) m = 2 2) m = 23 3) m = 4 4) m = 103

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171 5) m = 65 6) m = 34 7) m = 0 8) m = 3 9) m = 3 10) m = 2 11) m = 38 12) m = 13 13) Neither 14) Neither 15) Parallel 16) y = 2x +5 17) y = 35x+5 18) y = 4x 3 19) y = 2 20) y = x 1 21) y = 2x 11 22) x = 5 23) y = 2x+5 4.1 Solving S ystems of E quations by G raphing 1) Yes 2) No 3) No 4) ( 2, 3) 5) ( 2, 2) 6) (1, 1) 7) ( 1, 4) 8) Infinite number of solutions 9) No solution 4.2 Solving Systems by Substitution 1) ( 2, 0) 2) (0, 2) 3) (10, 1) 4) (3, 2) 5) Infinite number of solutions (x, x/3 + 4/3) 6) No solution 4.3 Solving Systems by Addition 1) (1, 2) 2) ( 5, 1) 3) ( 2, 3) 4) (0, 2) 5) (3, 6) 6) ( 6, 10) 7) 15 & 24 8) 25 & 10 4.5 Application: Value Problem s 1) 347 child tickets & 206 adult tickets are sold 2) 13 dimes and 29 quarters 3) He should invest $32,800 in stock & $7,200 in bonds. 4) $10 bills = 27 & $20 bills = 3 5) $12,000 should be invested at 5.25% & $13,000 should be invested at 4%. 4.6 Application: Mixture Problem s 1) 16 pounds of nuts & 4 pounds of chocolate chips 2) 3 pounds of peanuts & 2 pounds of cashews 3) 80% = 28 liters & 30% = 42 liters 4) 12% = 50 liters & 25% = 15 liters 5) 70% = 80 liters & 40 % = 160 liters

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172 5.1 Exponent Properties 1) 2) 8 3) 18 4) 2 5) 6 6) 4 7) 8) 6c 9) 3 10) 6 11) 11 12) 125 13) 216 14) 49 15) 64 16) 17) 18) 32 19) 18 20) 27 21) 243 22) 23) 24) 34 25) 9 5.2 Negative Exponents 1) 1 2) 1 3) 1 4) 1 5) 0 6) 2 7) 0 8) 9) 10) 11) 12) 13) 14) 15) 5.3 Scientific Notation 1) 5 . 75 10 2) 8 . 74 10 3) 6 . 43 10 4) 8 . 02 10 5) 25 . 4 6) 6190 7) . 464 8) 0 . 007 9) = 13 10) 3 . 09936 10 11) 3 . 55 10 12) 1 . 4 10 3 13) 1 . 56 10 14) 18 . 148148 10 15) 55 . 405405 10 5.4 Introduction to Polynomials 1) (a) 52 (b) 40 (c) 4 2) (a) monomial, degree 0 (b) polynomial, degree 3 (c) trinomial, degree 4 (d) binomial, degree 2 (e) monomial, degree 10 3) (a) 40y2 (b) 22pq3 4) 11y2 10y + 2 5) (a) 10m2 + 3mn 8n2 (b) 3b23ab (c) 5p3 6p2q+ pq2 6) 2m2 7m + 4 7) 3x2 – x + 4 8) 8z + 2

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173 5.5 Multiply Polynomials 1) 6p + 42 2) 32k2 + 16k 3) 18n3 21n2 4) n2 + 14n + 48 5) b2 2b 15 6) 4r2 24r 64 7) 49n2 36 8) 25x2 15xy + 2y2 9) 6r343r2 + 12r 35 10) 12n3 20n2 + 38n 20 11) 18x2 15x 12 12) 7x2 + 49x 70 13) x3 3x2 10x + 21 14) 10x + 2 15) P = 4w + 4; A = w2 + 2w 16) P = 6w 6; A = 2w2 3w 5.6 Multiply Special Products 1) x2 + 10x + 25 2) 4x24x + 1 3) 9x2 12xy + 4y2 4) x2 6xy + 9y2 5) x2 + x + 1/4 6) 3a2 24a + 48 7) 5w2 + 10wy 5y2 8) x2 9 9) d2 49 10) 9x2 1 11) 25x2 4y2 12) c2 121 13) x214) 5x2 5 15) 12x3 + 13x 16) x3 15x2 + 75x 125 17) 8x3 36x2 + 54x 27 5.7 Divid e Polynomials 1) 2) 3) 4) 3 n+ 2 n 5) 2a 1 6) 6x 10y 4x2y 7) y + 4 8) a – 7 9) 4x + 3 10) p + 3 ( ) 11) 3b 2 + ( ) 12) 16y2 + 12y + 9 13) (a23)

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174 6.1 FactoringGreatest Common Factor 1) 9 + 8b2 2) x 5 3) 5(9x2 5) 4) 1 + 2n2 5) 7(8 5p) 6) 10(5x 8y) 7) 7ab (1 5a) 8) 9x2y2(3y3 8x) 9) 3a2b( 1 + 2ab) 10) 4x3(2y2 + 1) 11) 5x2(1 + x + 3x2) 12) 8n5( 4n4 + 4n + 5) 13) 10(2x4 3x + 3) 14) 3(7p6 + 10p2 + 9) 15) 4(7m4 + 10m3 + 2) 16) 2x ( 5x3 + 10x + 6) 6.2 FactoringGrouping 1) (8r2 5) (5r 1) 2) (5x2 8) (7x 2) 3) (n2 3) (3n 2) 4) (2v2 1) (7v + 5) 5) (3b2 7) (5b + 7) 6) (6x2 + 5) (x 8) 7) (3x2 + 2) (x + 5) 8) (7p2 + 5) (4p + 3) 9) (7x2 4) (5x 4) 10) (7n2 5) (n + 3) 6.3 FactoringTrinomials where a=1 1) (p+ 9) (p+ 8) 2) (x 8) (x+ 9) 3) (n 8) (n 1) 4) (x 5) (x+ 6) 5) (x+1) (x 10) 6) (x+5) (x+ 8) 7) a(b+8) (b+ 4) 8) 2(b 10)(b 7) 9) (w 8) (w+1) 10) 4b(a 5) (a 2) 6.4 Factoring Trinomials where 1) (7x 6) (x 6) 2) (b+2) (7b+1) 3) (5a+7) (a 4) 4) (2x 1) (x 2) 5) (x+7) (2x+5) 6) (2b 3) (b+1) 7) (5k+3) (k+2) 8) (3x 5) (x 4) 9) (3x+2y) (x+5y) 10) (5x 7y) (x+7y) 11) 3(2x+1) (x 7) 12) 3(7k+6) (k 5) 13) 2(7x 2) (x 45) 14) (6x+5) (x+4) 15) (4k 1) (k 4) 16) (x+2y) (4x+y) 17) (4m+3n) (m 3n) 18) (x+3y) (4x+y) 19) 2(2x+7y) (3x+5y) 20) 4(6x y) (x 2y) 6.5 Factoring Special Products 1) (r+4) (r4) 2) (p+2) (p 2) 3) (3k+5) (3k 5) 4) 3(x+3) (x 3) 5) 4(x+3) (x 3) 6) 2(3a 5b) (3a 5b) 7) (a 1)2 8) (x+3)2 9) (x 3)2 10) (5a 1)2 11) (5a+3b)2 12) (2a 5b)2 13) 2(2x 3y)2 14) –(m 2) (m2+2m+4)

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175 15) (x 4) (x2+4x+16) 16) (5a 4) (25a2+20a+16) 17) (4x+3y) (16x212xy+9y2) 18) (a2+9) (a+3) (a 3) 19) –(z2+4) (z+2) (z 2) 20) (m2+9b2) (m+3b 6.6 Factoring Strategies 1) 2) 3) 2 + 4xy + 16y2) 4) 5) 6) 7) 8) 9) 10) 11) 12) 13) n (n +2) (n + 5) 14) 2 + 12x + 16) 15) x (5x + 2) 16) 17) 18) 2 19) 20) 21) 6.7 S ol ve by Factoring 1) 7, 2 2) 1, 4 3) 5,5 4) 2, 7 5) , 3 6) , 2 7) 4,0 8) 1,4 9) , 8 10) , 3 11) 4, 3 12) 8, 2 13) , 5 14) , 3 15) 4, 1 16) 7,7 17) , 8 18) , 6 7.1 FactoringGreatest Common Factor 1) 2) 3) 4) none 5) , 1 6) , 7) 8) 9) 10) ( ) 11) 12)

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176 13) 14) 15) 16) 17) 18) 19) 20) ( ) ( ) 21) 7.2 Multiply and Divide Rational Expressions 1) 2) 3) 4) 5) 6) ( ) ( ) 7) 8) ( ) 9) 10) 11) 12) 13) ( + 5 ) ( 2 ) 14) 15) 16) 17) ( ) 18) 19) 20) 2 ( 8 ) 21) 7.3 Least Common Denominator 1) 6 2) ( + 5 ) 3) ( 5 ) ( + 5 ) 4) ( + 1 ) ( + 2 ) ( + 3 ) 5) 4 ( 2 ) 6) ( + 1 ) ^ 2 ( + 2 ) 7) , 8) ( ) ( ) , ( ) ( ) 9) ( ) ( ) , ( ) ( ) 10) ( ) , ( ) , ( ) 11) ( ) , ( ) 12) ( ) ( ) , ( ) ( )

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177 13) ( )( ) , ( )( ) 14) , 15) ( ) ( ) ( ) , ( ) ( ) ( ) 16) ( ) ( ) ( ) , ( ) ( ) ( ) 17) ( ) , ( ) 18) ( ) ( ) , ( ) ( ) 19) ( )( ) , ( )( ) , ( )( ) 20) ( )( ) , ( )( ) 21) ( ) ( ) , ( ) ( ) , ( ) ( ) 7.4 Add and Subtract Rational Expressions 1) 2) 3) 4) ( ) ( ) ( ) 5) ( ) ( ) 6) ( )( ) 7) ( ) ( ) ( ) 8) 9) 10) 11) ( ) ( ) 12) 13) ( ) 14) ( ) ( ) ( ) 15) 16) 17) 18) ( ) ( ) ( ) ( ) 19) ( )( ) 20) 21) ( ) ( ) 7.5 Complex Fractions 1) 2) 2 3) 4) 5) 6)

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178 7) 8) 9) 10) 11) 12) 13) ( ) 14) ( ) 15) 16) 17) 18) 19) 20) 21) 7.7 Solve Rational Equations 1) 2) 7 3) 4) 5) 2 6) 2 , 3 7) 4 , 1 8) none 9) 7 10) 11) 12) 0 , 13) 8 14) 15) 16) 17) none 18) 43 19) 20) 8 8.1 Square Roots and 8.2 Higher Roots 1) 8 2) 5 3 3) 24 2 4) 3 2 5) 80 6 6) 7 2 7) 5 2 8) 5 5 9) 27 10) 4 3

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179 8.3 Adding Radicals 1) 5 3 2) 3 + 2 3) 3 3 4) 5 5) 5 2 6) 9 3 3 7 7) 6 2 8) 0 9) 17 50 + 3 3 10) 7 5 + 5 5 11) 5 3 6 4 8.4 Multiply Radicals 1) 2 2) 30 2 3) 5 3 4) 15 2 5) 42 5 6) 75 7) 4 10x 8) 11 + 10 11 9) 15 8 5 10) 31 11) 12) 103 + 15 35 13) 2 2 35 8.5 Divide and rationalize the denominator: 1) 2) 3) 4) 5) 3 + 2 6 6) 2 + 7 7) 8) 9) 10) ( ) 11) ( ) 12) 13) 27 + 10 6 14) 9.1 Quadratics Solving with Radicals 1) 2 2) 1,5 3) 2, 2 4) 5) 5 6) 3 7) 1 8) 9) 5 10) 2

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180 9.2 Solving with Exponents using the Square Root Property 1) x = 2, 10 2) a = 2 5 3 , 2 + 5 3 3) x = 2, 6 4) x = 5 2 2 , 5 + 2 2 5) h = , 1 6) x = 7 3 10, 7 + 3 10 7) x = , 8) a = 1 9) x = , 10) m = 2 9.3 Completing the Square 1) 25; (x+5)2 2) 16; (x 4)2 3) x = 1, 13 4) x = 3 + 10 , 3 10 5) x = 10, 8 6) x = 5, 1 7) x = 4 + 31 , 4 31 8) x = 3 + 7 2 , 3 7 2 9.4 Quadratics Quadratic Formula 1) 2 2) 1 3) 3, 4) 2, 3 5) 1, 6) 1, 7) 1+ 5 , 1 5 8) 1 9) 1, 10) No real number solution

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181 References: 1) College Algebra Senior Contributing Authors : Jay Abramson, Arizona State University Nicholas Belloit, Florida State College at Jacksonville https://openstax.org/details/books/college algebra 2) Intermediate Algebra 2e Senior Contributing Authors Lynn Marecek, Santa Ana College Andrea Honeycutt Mathis, Northeast Mississippi Community College https://openstax.org/details/books/intermediate algebra2e 3) Elementary Algebra 2e Senior Contributing Authors Lynn Marecek, Santa Ana College MaryAnne Anthony Smith, Formerly of Santa Ana College Andrea Honeycutt Mathis, Northeast Mississippi Community College https://openstax.org/details/books/elementary algebra 2e 4) Beginning and Intermediate Algebra Tyler Wallace http://wallace.ccfaculty.org/book/book.html 5) Introductory Algebra Student Workbook Fifth Edition Development Team Jenifer Bohart, Scottsdale Community College William Meacham, Scottsdale Community College Amy Volpe, Scottsdale Community College James Sousa, Phoenix College Judy Sutor, Scottsdale Community College Donna Gaudet, Scottsdale Community College https://myopenmaths3.s3.amazonaws.com/cfiles/09XWorkbook_Modules567_Fall2015_0 .pdf 6) Introductory Algebra Andrew Gloag Anne Gloag adapted by James Sousa http://www.opentextbookstore.com/sousa/CK12IntroAlg.pdf https://www.ck12.org/saythanks/