Beginning Algebra Part I Workbook

OERWorkbookFinal0106-Version4 Accessible_FINAL ( .docx )

OERWorkbookFinal0106-Version4 Accessible_FINAL ( .pdf )

PAGE 1

1 OTC Beginning Algebra Part I Palma Benko, David Burger, Yon Kim, Michael Walters, Akbar Afsheen, Laura Samuelsen, Deborah Poetsch

PAGE 2

2 Beginning Algebra Part I 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

PAGE 3

3 Please note this is an uncorrected working draft currently in revision for ADA Compatability 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 Attribution 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 Under 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 information 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 Collaborative 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. including health services, technology, energy, and global manufacturing and supply chain management as identified by the New

PAGE 4

4 Table of Contents Page Chapter 0: Arithmetic 0.1 Integers 5 0.2 Fractions 8 0.3 Order of Operations 11 0.4 Properties of Algebra 1 3 Chapter 1: Solving Linear Equations and Inequalities 1.1 One Step Equations 1 6 1.2 Two Step Equations 1 8 1.3 General Linear Equations 21 1.4 Solving with Fractions 2 3 1.5 Formulas 2 6 1.8 Application: Number/Geometry 30 1.9 Other Applications 3 6 1.10 Solve and Graph Inequalities 40 Chapter 2: Graphing 2.1 Graphing: Points and Lines 4 4 2.2 Slope 4 8 2.3 Slope Intercept Form 5 3 2.4 Point Slope Form 5 7 2.5 Parallel & Perpendicular Lines 61 Chapter 4: Systems of Equations 4.1 Solving Systems of Equations by Graphing 6 5 4.2 Solving Systems by Substitution 71 4.3 Solving Systems by Addition 7 4 4.5 Application: Value Problems 7 9 4.6 Application: Mixture Problems 8 3 Chapter 5: Polynomials 5.1 Exponent Properties 8 7 5.2 Negative Exponents 91 5.3 Scientific Notation 9 3 5.4 Introduction to Polynomials 9 6 5.5 Multiply Polynomials 9 9 5.6 Multiply Special Products 10 3 5.7 Divide Polynomials 10 7 Answer Key for Worksheets 11 2 References 11 9

PAGE 5

5 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 : 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, 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 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: Since both numbers are negative, we add their absolute values: The result will be negative: 0 1 2 3 4 5 6 7 8 1 2 3 4 5 6 7 8 Negative Positive

PAGE 6

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

PAGE 7

7 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) 9) 10) 11) 12) 13 ) 1 4 ) 15 ) 1 6 ) 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. ? 1 7 ) In Fargo it was 18°F, while in Tacoma it was 43°F. How much warmer was Tacoma? 1 8 Multiply or divide : 19 ) 2 0 ) 21 ) 22 ) 23 ) 24 ) ==========================================================================

PAGE 8

8 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 to an improper fraction. 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, 47÷6 = 7 remainder 5. So, there are 7 wholes, and 5 remaining pieces, giving the mixed number 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 =

PAGE 9

9 To multiply with mixed numbers, it is easiest to first convert the mixed numbers to improper fractions. Example E) Multiply and simplify Converting these to improper fractions first, and , so . 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: 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: Example G) Divide and simplify Rewriting the mixed numbers first as improper fractions, We find the reciprocal of and change this into a multiplication proble m 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 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 the two denominators, 4, and give both fractions that denominator. Then we add and simplify.

PAGE 10

10 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 Rewriting the fractional parts with a common denominator of 12: Adding the whole parts . Adding the fractional parts, . Now we combine these: ========================================================================== Worksheet: 0.2 Fractions Convert each mixed number to an improper fraction 1) 2) Convert each improper fraction to a mixed number 3) 4) Simplify to lowest terms 5) 6) 7) 8) Multiply and simplify 9) 10) 11) 12) 13) One dose of eyedrops is ounce. How many ounces are required for 40 doses? Divide and simplify 1 4 ) 1 5 ) 1 6 ) 1 7 ) 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 ) 25 ) 26) ==========================================================================

PAGE 11

11 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 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 the 3 is the exponent and it means the base number is to be multiplied by itself that number of times, so 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: We begin with the inside of the paren the s is , with the exponent: Still inside the parenthesis , we do the division: Inside the parenthesis , we add Now multiply Add 32 Example B) P E MD left to right AS left to right AS left to right Example C) P E MD left to right MD AS P: P arentheses E: E xponents and roots MD: M ultiplication and D ivision AS: A ddition and S ubtraction

PAGE 12

12 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 Our Solution Example E) P Start with ( ), inside ( ): exponents first: P Continue ( ): M Multiply ========================================================================== Worksheet : 0.3 Order of Op erations Simplify : 1) 2) 3) 4) 5) 6) 7) 8) 9) 10) 11) 12) 13 ) 14) 15 ) . (mean = add up all the numbers, then divide by how many numbers there are) ==========================================================================

PAGE 13

13 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: 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: terms are , 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: Evaluate algebraic expressions : Replace the variables with their numerical values and follow order of operations. Example A) Evaluate when and Replace p with 3 and q with 5: Evaluate parenthesis , add: Multipl y : 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: Combine like terms and and O ur solution

PAGE 14

14 Distributive Property : a(b + c) = ab + ac Example C ) Simplify: Multiply each term by 4 O ur Solution Example D ) Simplify: Our Solution Example E ) S implify : Our Solution Translate a lgebraic e xpressions Operation + × ÷ Algebraic Expression x + 28 k 12 (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

PAGE 15

15 Example F) Translate: the product of 8 and 5 less than a number o 5 less than a number: o o x 5 Translation: the product of 8 and 5 less than a number ========================================================================== Worksheet: 0.4 Properties of Algebra (Simplify Expressions) Evaluate each expression if , , and . 1. Evaluate each expression if , , , and . Combine like terms: 7 ) 8 ) 9 ) 10 ) Distribute: 1 1 ) 1 2 ) 1 3 ) 1 4 ) Simplify: 1 5 ) 1 6 ) 1 7 ) 1 8 ) Translate: w rite an algebraic expression to the given verbal expression. 1 9 . eight less than a number 20 . a number increased by seven 2 1 . the quotient of m and n 2 2 . a number squared 2 3 . the sum of 3 times a and b 2 4 . three times the sum of a and b 2 5 . seven more than the cube of a number 2 6 . one half the product of x and y 2 7 . the product of twice a and b 2 8 . twice the product of a and b 2 9 . two less than five times a number 30 . twice a number increased by three times the number 3 1 . the cube of a plus b 3 2 . the cube of the sum of a and b ==========================================================================

PAGE 16

16 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: Note: Whatever you do to one side, you MUST do to the other side (keep it balanced). When solving equations, eliminate double signs. As a general rule, replace ( ) with and ( with . Example B) Solve: Solve using multiplication and division. Example C) Solve: Get the variable by itself. Right now, 16 is being added to it. Undo the addition by subtracting 16 from both sides. Answer. Undo the subtraction by adding 3 to both sides. Answer. 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.

PAGE 17

17 R emember: Whatever you do to one side, you MUST do to the other side (keep it balanced). Example D) Solve : Since 4 is dividing x, multiply both sides by 4 to clear the fraction. The fours will cancel each other out. simplifies to 1x Example E) Solve: 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. So, 7 is the solution to the ========================================================================== Worksheet: 1.1 Solving Linear Equations One Step Equations. 1) 2) 3) 4) 5) 6) 7) 8) 9) 10) 11) 12) 13) 14) 15) 16) ======================================================================== Multiplication Property of Equality Multiplying the same value from both sides of the equation.

PAGE 18

18 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 two step 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 : Step 1: Find the term with the variable Step 2: What is happening to that term? 5 is being added to 2 x Step 3: Do the opposite operation to both sides of the equation. Subtraction is the opposite operation of addition, so subtract 5 from both sides. Step 4: Find the variable. Step 5: What is happening to the variable? x is being multiplied 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. Division is the opposite operation of multiplication, so divide both sides by 2. See that x = 4. Check: Substitute the answer back into the original equation. Since 13 is 13, our answer is correct. Example B) S olve: Start by focusing on the plus 7 Subtract 7 from both sides Now focus on the multiplication by 5 Divide both sides by 5 Our Solution!

PAGE 19

19 Example C ) S olve: Start by focusing on the positive 4 ( Subtract 4 from both sides N egative (subtraction) stays on the Divide both sides by 2 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 min) Solution : The repair job was completed at 9:30 + 1:45 = 11:15AM ==========================================================================

PAGE 20

20 Worksheet: 1.2 Solving Linear Equations Two Step Equations. 1) 2) 3 ) 4 ) 5 ) 6 ) 7 ) 8 ) 9 ) 10 ) 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? 1 4 ) 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. 1 5 ) 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 costs $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. ==========================================================================

PAGE 21

21 1.3 General Linear Equations Multi Steps Equations Learning Objectives : In this section, you will: Solve a multi step equation by combining like terms Solve a multi step 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 7) Check the solution. Example A) Solve: 3x + 9 = 6x 27 Notice the variable on both sides, 3x is smaller Add 3x to both sides Focus on the subtraction by 27 Add 27 to both sides Focus on the multiplication by 9 Divide both sides by 9 Our Solution Check: True, so x = 4 is the solution. Example B ) Solve: Distribute 4 through parenthesis Focus on the subtraction first Add 24 to both sides Now focus on the multiply by 8 Divide both sides by 8 Our Solution! Check: True, so x = 5 is the solution.

PAGE 22

22 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: Distribute 3 through parenthesis Notice the variable on both sides Subtract 6x from both sides 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: Distribute 2 through parenthesis Combine like terms Notice the variable is on both sides Subtract 2x from both sides 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) 2) 3) 4) 5) 6) 7) 8) 9) 10)

PAGE 23

23 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 Denominator . 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 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. Use the Distributive Property. Simplify and notice, no more fractions! Step 3. Solve using the General Strategy for Solving Linear Equations. To isolate the variable term, subtract 10. Simplify. Check:

PAGE 24

24 Example B ) Solve: LCD = 12 , multiply each term by 12 Reduce each 12 with denominators Multiply out each term Focus on subtraction by 42 Add 42 to both sides 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: L CD = 6, multiply each term by 6 Reduce 6 with each denominator Multiply out each term Notice variable on both sides Subtract 42 from both sides Focus on addition of 1 Subtract 1 from both sides Focus on multiplication of 5 Divide both sides by 5 Our Solution

PAGE 25

25 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: Solution Distribute. Simplify. Multiply by the LCD, 4 . Distribute. Simplify. Collect the variables to the left Simplify. Collect the constraints to the right Simplify. ========================================================================== Worksheet: 1.4 Solving with Fractions Solve each equation . Make sure any fractional solutions are in simplest form. 1) 2) 3) 4) 5) 6) 7) 8) 9) 10)

PAGE 26

26 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 equa tion, the second is a formula that we are solving for x. Example A ) Solve for x: In both problems, x is multiplied by something To isolate the x, we divide by 3 or w. Our Solution We use the same process to solve 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.

PAGE 27

27 Example C ) Solve : Solution We will isolate y on one side of the equation. Subtract 8 x from both sides to isolate the term with y. Simplify. 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 ) 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.

PAGE 28

28 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. Remove the fraction on the right. Simplify Divide both sides by 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. ==========================================================================

PAGE 29

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

PAGE 30

30 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. etc. 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? Subtraction is built backwards, multiply the unknown by 5 Is translates to equals Add 28 to both sides The variable is multiplied by 5 Divide both sides by 5 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 First, addition is built backwards Then, subtraction is also built backwards Is between the parts tells us they must be equal Subtract so variable is all on one side

PAGE 31

31 Now we have a two step equation Add 10 to both sides 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 Together they cost 444, so we add. Replace S and L with labeled values Parentheses are not needed, combine like terms Divide both sides by 3 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. ==========================================================================

PAGE 32

32 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? 4. A certain number added twice to itself equals 96. What is the number? 6. Sixty more than nine times 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? ==========================================================================

PAGE 33

33 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 of 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

PAGE 34

34 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. With nothing given about the first we make that x The second is double the first, The third is 40 less than the first All three angles add to 180 Replace F, S, and T with the labeled values. Here the parenthesis is not needed. Combine like terms, Add 40 to both sides The variable is multiplied by 4 Divide both sides by 4 Our solution for x Replace x with 55 in the original list of angles Our angles are 55, 110, and 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 often 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. The dimensions are L = 15 ft and W=12ft .

PAGE 35

35 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 Width is five less than two times the length The formula for perimeter of a rectangle Replace P, L, and W with labeled values Distribute through parenthesis Combine like terms Add 10 to both sides The variable is multiplied by 6 Divide both sides by 6 Our solution for x Length 9 Replace x with 9 in the original list of sides 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 twice 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. ==========================================================================

PAGE 36

36 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 equation 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. David so he is x, Carmen then is x+12 S ubtract 5 from now column to get the change The sum of their ages will be 29. So, we add C and D Replace C and D with the change cells. Remove parenthesis Combine like terms and 7+5 Subtract 2 from both sides Age Now 5 Carmen David Age Now 5 Carmen David Age Now 5 Carmen David

PAGE 37

37 Notice x is multiplied by 2 Divide both sides by 2 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 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 mark up 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 David

PAGE 38

38 III. Commission problems Commission is paid to an employee as an incentive to sell more. A commission is generally a percentage of sales. Commission percentage and should be converted to a decimal) is given by: Example D ) Solve: The Grey family's house was sold for $ 2 00,000. 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( 2 00,000 ) = $1 0 ,000 The real estate agent will get $1 0 ,000 $ 2 00,000 $1 0 ,000 = $ 190 ,000 The family will get $ 190 ,000 after they pay their real estate agent. Discount formula 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.

PAGE 39

39 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% commission 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 commission 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. ==========================================================================

PAGE 40

40 1.10 Solve and Graph Inequalities In this section, you will: Solve for the solutions to linear inequalities. 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 is 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 the 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 ne gative infinity, we will always use a curved bracket for that value. Example A) Graph the inequality and give the interval notation 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)

PAGE 41

41 Example B) Graph the inequality and give the interval notation start at 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 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 : Subtract 5 from both sides Divide both sides by 2 Divide by a negative flip symbol! Graph, starting at 3, going down with ] for less than or equal to ADA: Number line shaded: 3 , ] at 3 ; Interval notation: ( , 3]

PAGE 42

42 Example E) Solve and give the result in interval notation : Distribute Combine like terms Move variable to one side Subtract 10x from both sides Add 20 to both sides Divide both sides by 2 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. ==========================================================================

PAGE 43

43 Worksheet : 1.10 Solve and Graph Inequalities Draw a graph for each inequality and give interval notation. 1) 2) 3) 4) 5) 6) 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 ) 11 ) 12 ) 1 3 ) 1 4 ) 1 5 ) 1 6 ) 1 7 ) ==========================================================================

PAGE 44

44 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 Pair s 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. x axis: the horizontal number line. y axis: 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 where 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

PAGE 45

45 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 following points on the coordinate grid . 1 3 . S ( 6, 3 ) 1 4 . T ( 2 , 4 ) 1 5 . U ( 16. V (4.5, 3.2) 17. W (0, 3) 18. X ( Identify the quadrant in which the point is located (ex 16 19) . 19. Point B in graph ________ 20. Point E in graph ___________ 21. Point ( 20, 50) ________ 22. Point (3.5, 100) ___________

PAGE 46

46 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 2. Make the graph a. The straight line shows all possible solutions to the equation calculations in step #1 and the plotting of ordered pairs Example B ) Graph: Use Use Use Example C ) Graph: Use Use Use T Table x y 0 2 1 1 2 4 T Table x y 0 5 1 3 2 1

PAGE 47

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

PAGE 48

48 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 :

PAGE 49

49 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.Substitu t 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

PAGE 50

50 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 . T here is a way to find the slope without graphing. We will use two points: ( x 1 , y 1 ) to identify the first point and ( x 2 , y 2 ) to identify the second point. Slope of a Line The slope m of the line containing the points ( x 1 , y 1 ) and ( x 2 , y 2 ) is given by Example B) Find the slope of the line through (4, 3) and (2, 2). Step 1 : Label your points Let ( x 1 , y 1 ) be (4, 3) and ( x 2 , y 2 ) be (2, 2) Step 2 : Use the slope formula and substitute the value Note: If we let ( x 1 , y 1 ) be (2, 2) and ( x 2 , y 2 ) be (4, 3), then we get the same result.

PAGE 51

51 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 X AND Y INTERCEPTS 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: 2x 3y = 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 ) ==========================================================================

PAGE 52

52 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 x and 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 x and y intercepts from the equations: 10. 11. 12. 13. 14.

PAGE 53

53 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 equation. 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 ) points (0,1) and (1,3). Find the rise and run. 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)

PAGE 54

54 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. Write the equation of the line. Identify the slope. Identify the y intercept. y intercept is (0,5) Example C) Identify the slope and y intercept of the line with equation 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 using its slope and y intercept. Solution: Step 1. Fine the slope intercept form of the equation. This equation is in slope intercept form. Step 2. Identify the slope and y intercept. Use Find the slope. Find the y intercept. Solve for y . Subtract x from each side. Divide both sides by 2. Simplify. Simplify. Write the slope intercept form of the equation of the line. Write the equation of the line. Identify the slope. Identify the y intercept. y intercept is (0,3)

PAGE 55

55 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. 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. ==========================================================================

PAGE 56

56 Worksheet : 2.3 Graphing: Slope Intercept Form Use the graph s to find the slope and y intercept of the line s . 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) 5 ) 6 ) Write the slope intercept 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) 12) 13) 14)

PAGE 57

57 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 requires 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 (x 1 , y 1 ), 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, ( x 1 , y 1 ), (x, y) P lug in the values Multiply both side by 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 . If the slope is not given, then you must find the slope to use the formula.

PAGE 58

58 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 Step 2. Identify the point The point is given. Step 3. Substitute the values into the point slope form, Simplify. Step 4. Write the equation in slope intercept form. 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. Identify the point. Substitute the values into Simplify. It is in y form but could be written Write in slope intercept form : Write in point slope form: 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

PAGE 59

59 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. 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. Step 2 . Choose one point. Choose either point. Step 3 . Substitute the values into the point slope form, Simplify. Step 4. Write the equation in slope intercept form. 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 and y intercept (0, b) slope intercept Slope and a point : P (x1, y1) point slope Two points : P1 (x1, y1); P2 (x2, y2) point slope ==========================================================================

PAGE 60

60 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 the 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 ). ==========================================================================

PAGE 61

61 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 To find slope we will put equation in slope intercept form Subtract 32 from both sides Put x term first Divide each term by 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.

PAGE 62

62 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 ). Step 2: Identify the slope of line 2. The slope of this equation will have to be found. To find the slope, arrange the equation to slope intercept form by solving for y In this form the slope is given (coefficient of ) Step 3: Compare 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 : 2. Find the slope of the parallel or perpendicular line. 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 Step 3. Ident ify the point. The given point is ( 2,1) Step 4. Substitute the values into the point slope form, Simplify. Step 5. Find the slope of the given line. The line is in slope intercept form,

PAGE 63

63 Example E ) 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, 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) Step 4. Substitute the values into the point slope form, Simplify. 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 . ==========================================================================

PAGE 64

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

PAGE 65

65 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 . 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 . 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 below 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 ? Identify x and y from the ordered pair Plug these values into each equation First equation Evaluate True

PAGE 66

66 Second equation, evaluate 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 ? Identify x and y from the ordered pair Plug these values into each equation First equation Evaluate 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 equation s . 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 Step 1. Graph the first equation To graph the first line, write the equation in slope intercept form.

PAGE 67

67 Step 2. Graph the second equation on the same rectangular coordinate system To graph the second line, use intercepts. (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). The solution is (4, 1) Example D ) Solve the System of Linear Equations by Graphing 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. Find the slope and y intercept of the first equation. Graph the two lines.

PAGE 68

68 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 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 Solution: Identify slope and y intercept of each equation Fi r st: Now we can graph both equations on the same plane Second 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 would have known there was no solution even without having to graph lines.

PAGE 69

69 Example F) Solve the System of Linear Equations by Graphing Solution: Subtract x terms Put x terms first Divide by coefficient of y Identify the slopes and y intercepts First: Now we can graph both equations t ogether Second: 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 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 Linear System : Graph Number of solutions 2 intersecting lines 1 Parallel lines None Same line Infinitely many ==========================================================================

PAGE 70

70 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 ? 2) Is ( 3, 1) is the solution to the system ? 3) Is ( 2, 0) is the solution to the system ? Solve the following systems of linear equations using graphing method : 4) 5) 6) 7) 8) 9)

PAGE 71

71 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 . 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: (1) (2) Solution: Since the first equation has y with a coefficient of 1, we will solve the first equation for y (3) ( 2 ) We will then substitute what we have for y equation (3) into equation ( 2 ) and solve Substitution Distribute Collect like terms Isolate the variable Divide away the coefficient Now that we know what x is, we can substitute it into equation (3) to get y So, the so lution is (2 , 3). Remember we can always check our solution by plugging it into

PAGE 72

72 the original system. Example B) Solve the system by substitution: ( 1 ) ( 2 ) Solution: Since the second equation (2) is already solved for x , we can go straight to plugging the solved form into equation ( 1 ) Substitution Collect like terms Isolate the variable Divide away the coefficient Now that we know what y is, we can substitute it into equation ( 2 ) to get x 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: ( 1 ) ( 2 ) Solution: Since the first equation has x with a coefficient of 1, we will solve the first equation for x ( 3 ) We can now substitute equation ( 3 ) into equation ( 2 ) Substitution Distribute 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 .

PAGE 73

73 Worksheet: 4.2 Solving Systems by Substitution Find the solutions to the systems: 1) 2) 3) 4) 5) 6)

PAGE 74

74 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 variable 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: (1) (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 coefficients of one variable be opposites, so that we can add the equations together and eliminate that variable. Here it is ea sy to eliminate y 1 as coefficients : (1) (2) Add (1) and (2) eliminate and we have one equation with one variable . Divide both sides by 5 So, the solution is (1, 2)

PAGE 75

75 Example B ) Solve the system by the addition method : 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. 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 by 2. Step 3. Add the equations resulting from Step 2 to eliminate one variable. We add the constants. 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 into the second equation, Then solve for y. Step 6. Write the solution as an ordered pair. Write it as (x,y). 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! The solution is (4, 1) .

PAGE 76

76 Example C ) Note: addition and elimination are the same methods . Solve the system by elimination: 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 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 Simplify Add the two equations to eliminate y. Solve for x. Substitute x = 0 into one of the original equations. Solve for y. 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 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.

PAGE 77

77 Example D) Solve the system by the addition method : 3x + 4y = 12 Solution Write the second equation in standard form. Clear the fractions by multiplying the second equation by 4. Simplify. To eliminate a variable, we multiply the second equation by 1. Simplify and add. 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:

PAGE 78

78 Worksheet: 4. 3 Solving Systems by Addition Find the solutions to the systems: 1. 2. 3. 4. 5. 6. 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.

PAGE 79

79 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 item s 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

PAGE 80

80 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. 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. We can solve this system by any method we like, but since the second equation is solved for d already, we will use substitution Substitution Distribute Collect like terms Isolate the variable Divide away the coefficient Now that we know how many nickels Natasha has, we can substitute it to find the number of dimes. This means for our solution; Natasha has 63 dimes and 36 nickels.

PAGE 81

81 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. Again, we will use substitution since both coefficients in the first equation are one. We will solve for c We can now substitute the expression for c into the second equation to find a . Substitution Distribute Collect like terms Isolate the variable Divide away the coefficient Now that we know how adult tickets were sold, we can substitute to find the number of child tickets sold. This means for our solution, there were 42 adult tickets sold and 105 child tickets sold.

PAGE 82

82 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 the 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?

PAGE 83

83 4.6 Application: Mix ture 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 contains 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 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.

PAGE 84

84 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. Amount of each item 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. We can solve this system by any method we like, but since the second equation is solved for d already, we will use substitution Substitution Distribute Collect like terms Isolate the variable Divide away the coefficient Now that we know how many nickels Natasha has, we can substitute it to find the number of dimes. This means for our solution; Natasha has 63 dimes and 36 nickels.

PAGE 85

85 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 will 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. Again, we will use substitution since both coefficients in the first equation are one. We will solve for x . We can now substitute the expression for x into the second equation to find y . Substitution Distribute Collect like terms Isolate the variable Divide away the coefficient Now that we know how much of the 75% solution to use, we can determine the amount of 90% solution. This means for our solution; we need to use 120 liters of the 75% solution with 240 liters of the 90% solution.

PAGE 86

86 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% solution 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?

PAGE 87

87 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 4 2 , the number 4 is the base and 2 is the exponent and called an exponential expression. Exponential expression (notation) In the expression a m , the exponent 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 is a constant and is a whole number. Example A) Evaluate : -------3 factors (Order of operations, multiply left to right) (Order of operations, exponent first) (Order of operations, exponent first ) Product Rule of Exponents: When multiply like bases, keep the base, and add the powers. Example B ) Simplify : Same base, add the exponents Our Solution Example C ) Simplify: Multiply 2.5, add exponents on x, y, and z Our Solution

PAGE 88

88 Example D ) Simplify: 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 Our Solution Example F ) Simplify: Subtract exponents on a, b, and c Our Solution Example G ) Simplify: 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: Put the exponent of 3 on each factor, multiplying powers Evaluate Our Solution Example K ) Simplify: a) b)

PAGE 89

89 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 4 th 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: Parenthesis are already simplified, next use power rules Using product rule, add exponents and multiply numbers Our Solution Example P ) Simplify: Use power rule in denominator Use quotient rule Our Solution ========================================================================== Rule of Exponents Product Rule of Exponents Quotient Rule of Exponents Power of a Power Rule of Exponents Power of a Product Rule of Exponents Power of a Quotient Rule of Exponents

PAGE 90

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

PAGE 91

91 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: Any number or expression raised to the zero power will always be 1. Example A F ) Evaluate: A) B) C) D) E) F) Rules of Negative Exponents: 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: Example H ) Simplify: **** All final answers should be written with positive powers.**** ==========================================================================

PAGE 92

92 Worksheet : 5.2 Negative Exponent 1. 2. 3. 4. 5. 6. 7. 8. 9. 10. 11. 1 2 . 1 3 . 1 4 . 1 5 . ==========================================================================

PAGE 93

93 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 a | < 10 and 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. in the same way 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) 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 to get back the original number. 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) 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 to get back the original number. multiplication. Example C) 10,400,000 in scientific notation equals Example D) .00204 in scientific notation equals

PAGE 94

94 Converting from Scientific Notation to Standard Notation (Decimal) We can also turn a number notated scientifically into a standard notation decimal number by reversing this process : Example E ) Convert into a decimal. = 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 into a decimal. = .00000054 C onvert this to decimal by moving the decimal point 7 places to the left (negative exponent means we divide). Example G ) = mov e the decimal point 4 places to the right = 63000 Example H ) = mov e the decimal point 3 places to the left = .00932 Operations on scientific numbers: Example I ) Multiply scientific numbers , find result in scientific notation : Multiply numbers 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.) Divide numbers Change this number into scientific notation Use product and quotient rule, using from the conversion Be careful with signs: Our Solution

PAGE 95

95 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 . 6 . 7 . 8 . 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 . 1 0 . One light year is approximately miles. Use scientific notation to express this distance in feet (Hint: 5,280 feet = 1 mile). 1 1 . 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. 13. 14. 15.

PAGE 96

96 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 Monomial Binomial Trinomial 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 3 Trinomial 2,1,0 2 1 Monomial 6 6 5 Polynomial 5,3,2,1,0 5

PAGE 97

97 Evaluate polynomials : replace the variable with the given number value and evaluate the polynomial. Example C ) Evaluate the trinomial : when x = 4 w hen x = 4 Replace variable x with 4 Exponents first Multiplication (we can do all terms at once) Add Our Solution Add/subtract polynomials: we combine like terms (add/subtract coefficients and keep variable) Example D ) Add and ) Combine like terms and Our Solution Example E ) Subtract from Distribute negative through second part Combine like terms and 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 ==========================================================================

PAGE 98

98 Worksheet : 5.4 Introductions to P olynomial s 1) Evaluate: For the polynomial , 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) c) d) e) 3) Add or subtract: a) b) 4) Find the sum: 5) Add or subtract: a) b) c) 6) Subtract: 7) Subtract from 8) Find the diffrence of and =========================================================================

PAGE 99

99 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: Distribute. Multiply. Multiply a b inomial by a b inomial : Two methods: distribute or FOIL Example B ) Multiply using the distributive property and FOIL Method: FOIL Distributive Property FOIL F O I L F O I L

PAGE 100

100 Example C ) Multiply using distributive property Distribute. Distribute again. Combine like terms. 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: Multiply the First. F O I L Multiply the Outer. F O I L Multiply the Inner. F O I L Multiply the Last. F O I L Combine like terms there are none.

PAGE 101

101 Multiply a polynomial by a polynomial : use distributive property Example E ) Multiply : Distribute. Multiply. Combine like terms Example F ) Multiply : Distribute 2x and 5 Multiply out each term Combine like terms Our Solution Example G ) Us e three methods to multiply binomial s : Distribute FOIL Rows ==========================================================================

PAGE 102

102 Worksheet : 5.5 Multiply Polynomials Find each product: 1. 2. 3. 4. 5. 6. 7. 8. 9. 10. 11. 12. 13. 14. Find the formula s for the perimeter and area of a rectangle where the and the 15. Find the formula s for the perimeter and area 16. Find the formula s for the perimeter and area of ==========================================================================

PAGE 103

103 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 What does this mean? It means to multiply by itself. Then, using FOIL, we get: Combining like terms gives: 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. Example A) Square the binomial using distributive property Squared is same as multiplying by itself Distribute Distribute again through final parenthesis Combine like terms Our Solution

PAGE 104

104 Example B ) Square the binomial ( x 5 ) 2 using perfect square formula above Recognize perfect square Square the first Twice the product Square the last Our Solution Example C ) Square the binomial using perfect square formula Recognize perfect square Square the first Twice the product Square the last 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

PAGE 105

105 Example E ) Multiply R ecognize sum and difference Square both, put subtraction between. Our Solution Example F ) Multiply Recognize sum and difference Square both, put subtraction between. Our Solution Example G ) Review the difference between the three problems : Binomial Squares Product of Conjugates 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. ==========================================================================

PAGE 106

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

PAGE 107

107 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 re discover the property by looking at some examples. Consider And What do they mean? Use the Equivalent Fractions Property. Simplify. QUOTIENT PROPERTY FOR EXPONENTS If a is a real number, a 0, and m and n are whole numbers, then Example A) Divide or find the quotient of and Rewrite as a fraction. Use fraction multiplication. Simplify and use the Quotient Property. Example B ) Divide or find the quotient of and Rewrite as a fraction Use fraction multiplication Simplify

PAGE 108

108 Divide a polynomial by a monomial : divide each term in the numerator by the monomial in denominator. Example C ) Find the quotient: Rewrite as a fraction. Separate the terms Simplify Example D ) Divide: Divide each term in the numerator by Reduce each fraction, subtracting exponents 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 Our Solution

PAGE 109

109 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 (x 2 + 9x + 20) by (x + 5) Note: realize that we cannot follow processes earlier , since divisor is not monomial . Write it as a long division problem. Be sure the dividend is in standard form. Divide need to multiply x by to get " x Put the answer, x, in the quotient over the x term. Multiply x times . Line up the like terms under the dividend. x Subtract from You may find it easier to change the signs and then add. Then bring down the last term, 20. x Put the answer, 4, in the quotient over the constant term. x Multiply 4 times x + 4 Subtract from x + 4 0 Check : Multiply the quotient by the divisor. You should get the divided.

PAGE 110

110 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: Write it as a long division problem. Be sure the dividend is in standard form with placeholders for missing terms. Divide by x. Put the answer, , in the quotient over the term. Multiply times . Line up the like terms Subtract and then bring down the next term. Divide by x. Put the answer, , in the quotient over the term. Multiply times x+1. Line up the like terms Subtract and bring down the next term. Divide 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. Divide by x. Put the answer, 1, in the quotient over the x term. Multiply 1 ties x+1. Line up the like terms Subtract and bring down the next term. Write the remainder as a fraction with the divisor as the denominator To check, multiply The result should be

PAGE 111

111 Example H ) Find the quotient: 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 : 2) Divide a polynomial by a monomial: 4) Divide a polynomial by a polynomial (use long division): 7) 8) 9) 10) 11) 12) 13) ===================================================================== To check, multiply The result should be

PAGE 112

112 Beginning Algebra Answer Key for the Worksheets 0.1 Integers 1) 2) 3) 4) 5) 6) 7) 8) 9) 10) 11) 12) 13) 14) 15) 16) 17) 18) 19) 20) 21) 22) 23) 24) 0.2 Fractions 1) 2) 3) 4) 5) ½ 6) 7) 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) 0.4 Properties of Algebra (Simplify, Evaluate, Translate Expressions) 1) 2) 3) 4) 5) 6) 7) 8) 9) 10) 11) 12) 13) 14) 15) 16) 17) 18) 19) 20) 21) 22) 23) 24) 25) 26) 27) 28) 29) 30) 31) 32)

PAGE 113

113 1.1 Solving Linear Equations One Step Equations 1) 2) 3) 4) 5) 6) 7) 8) 9) 10) 11) 12) 13) 14) 15) 16) 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) 2) 3) 4) 5) 6) 7) 8) 9) 10) 11) 1.5 Formulas 1) 2) 3) 4) 5) 6) 7) 8) 9) 10) 11) 12) 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

PAGE 114

114 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) 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)

PAGE 115

115 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; y int: (0, 4) 2) Slope: 1; y int: (0, 5) 3) Slope: 3/5; y int: (0, 1) 4) Slope: 53; y int: (0, 6) 5) Slope: 4/5; y int: (0, 8/5) 6) Slope: 4; y int: (0, 9) 7) y = 2x + 5 8) y = x 4 9) y = 3x 1 10) y = 13 x + 1 11) 1 3 ) 2.4 Point Slope Form 1) 2) 3) 4) 5) x = 6 6) 7) 8) 9) 10) y = 3 2.5 Parallel & Perpendicular Lines 1) m = 2 2) m = 23 3) m = 4 4) m = 103 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

PAGE 116

116 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 solution s 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: Mix ture 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 % = 8 0 liters & 40 % = 160 liters 5.1 Exponent Properties 1) 2) 3) 4) 5) 6) 7) 8) 6c 9) 10) 11) 12) 13) 14) 15) 16) 17) 18) 19) 20) 21) 22) 23) 24) 25)

PAGE 117

117 5.2 Negative Exponents 1) 2) 3) 4) 5) 6) 7) 8) 9) 10) 11) 12) 13) 14) 15) 5.3 Scientific Notation 1) 2) 3) 4) 5) 6) 7) 8) 9) 10) 11) 12) 13) 14) 15) 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) 40y 2 (b) 22pq 3 4) 11y 2 10y + 2 5) (a) 10m 2 + 3mn 8n 2 (b) 3b 2 3ab (c) 5p 3 6p 2 q+ pq 2 6) 2m 2 7m + 4 7) 3x 2 x + 4 8) 8z + 2 5.5 Multiply Polynomials 1) 6p + 42 2) 32k 2 + 16k 3) 18n 3 21n 2 4) n 2 + 14n + 48 5) b 2 2b 15 6) 4r 2 24r 64 7) 49n 2 36 8) 25x 2 15xy + 2y 2 9) 6r 3 43r 2 + 12r 35 10) 12n 3 20n 2 + 38n 20 11) 18x 2 15x 12 12) 7x 2 + 49x 70 13) x 3 3x 2 10x + 21 14) 10x + 2 15) P = 4w + 4; A = w 2 + 2w 16) P = 6w 6; A = 2w 2 3w

PAGE 118

118 5.6 Multiply Special Products 18) x 2 + 10x + 25 19) 4x 2 4x + 1 20) 9x 2 12xy + 4y 2 21) x 2 6xy + 9y 2 22) x 2 + x + 1/4 23) 3a 2 24a + 48 24) 5w 2 + 10wy 5y 2 25) x 2 9 26) d 2 49 27) 9x 2 1 28) 25x 2 4y 2 29) c 2 121 30) x 2 31) 5x 2 5 32) 12x 3 + 13x 33) x 3 15x 2 + 75x 125 34) 8x 3 36x 2 + 54x 27 5.7 Divid e Polynomials 1) 2) 3) 4) 5) 6) 6x 10y 4x 2 y 7) y + 4 8) a 7 9) 4x + 3 10) 11) 12) 16y 2 + 12y + 9 13) (a 2 3)

PAGE 119

119 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 algebra 2e 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_Fall201 5_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/