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SSIS Math 8 12-13

Gary Bertoia

Jen Kershaw

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AUTHORS

Gary Bertoia

Jen Kershaw

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Contents www.ck12.org

Contents

1 Single Variable Equations 1

1.1 Simplifying Algebraic Expressions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2

1.2 Expressions and the Distributive Property . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6

1.3 Translate Verbal Phrases into Variable Expressions. . . . . . . . . . . . . . . . . . . . . . . . . 13

1.4 One-Step Equations and Inverse Operations . . . . . . . . . . . . . . . . . . . . . . . . . . . . 17

1.5 Solve Equations Involving Inverse Properties of Addition and Multiplication . . . . . . . . . . . 21

1.6 Solve Equations Involving Inverse Properties of Addition and Division . . . . . . . . . . . . . . 27

1.7 Solve Equations Involving Inverse Properties of Subtraction and Multiplication . . . . . . . . . 32

1.8 Solve Equations Involving Inverse Properties of Subtraction and Division. . . . . . . . . . . . . 36

1.9 Solve Equations Involving Combining Like Terms . . . . . . . . . . . . . . . . . . . . . . . . . 41

1.10 Solve Equations with the Distributive Property . . . . . . . . . . . . . . . . . . . . . . . . . . . 48

1.11 Solve Equations with the Distributive Property and Combining Like Terms . . . . . . . . . . . . 52

1.12 Section Break . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 57

1.13 Solve Equations with a Variable on Both Sides . . . . . . . . . . . . . . . . . . . . . . . . . . . 59

1.14 Solve Multi-Step Equations Involving Decimals . . . . . . . . . . . . . . . . . . . . . . . . . . 65

1.15 Solve Multi-Step Equations Involving Fractions . . . . . . . . . . . . . . . . . . . . . . . . . . 71

1.16 Applications of One-Step Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 77

1.17 Solve Multi-Step Equations Involving Rational Numbers . . . . . . . . . . . . . . . . . . . . . 81

1.18 Problem-Solving Models . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 88

1.19 Comparison of Problem-Solving Models . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 97

1.20 Distributive Property for Multi-Step Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . 1061.21 Solving Real-World Problems Using Multi-Step Equations . . . . . . . . . . . . . . . . . . . . 110

2 Solving Inequalities 114

2.1 Inequalities that Describe Patterns . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 115

2.2 Inequalities with Addition and Subtraction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 119

2.3 Inequalities with Multiplication and Division. . . . . . . . . . . . . . . . . . . . . . . . . . . . 122

2.4 Solve Inequalities by Using the Distributive Property . . . . . . . . . . . . . . . . . . . . . . . 127

2.5 Multi-Step Inequalities . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 132

2.6 Applications with Inequalities. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 137

2.7 Inequality Expressions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 140

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www.ck12.org Chapter 1. Single Variable Equations

CHAPTER 1 Single Variable EquationsChapter Outline

1.1 SIMPLIFYING ALGEBRAICEXPRESSIONS

1.2 EXPRESSIONS AND THE D ISTRIBUTIVEPROPERTY

1.3 TRANSLATEV ERBAL P HRASES INTO VARIABLEE XPRESSIONS

1.4 ONE-S TE PEQUATIONS AND INVERSE OPERATIONS

1.5 SOLVEE QUATIONS I NVOLVINGI NVERSE P ROPERTIES OFA DDITION ANDM UL -

TIPLICATION

1.6 SOLVEEQUATIONS INVOLVING I NVERSE P ROPERTIES OF A DDITION ANDD IV I-

SION

1.7 SOLVE EQUATIONS INVOLVING INVERSE PROPERTIES OF SUBTRACTION AND

MULTIPLICATION1.8 SOLVE EQUATIONS INVOLVING INVERSE PROPERTIES OF SUBTRACTION AND

DIVISION

1.9 SOLVEEQUATIONS INVOLVING C OMBININGLIK ETERMS

1.10 SOLVEEQUATIONS WITH THEDISTRIBUTIVEPROPERTY

1.11 SOLVEE QUATIONS WITH THED ISTRIBUTIVEP ROPERTY ANDC OMBININGL IK E

TERMS

1.12 SECTION BREAK

1.13 SOLVEEQUATIONS WITH A VARIABLE ONBOTH SIDES

1.14 SOLVEMULTI-STE P E QUATIONS INVOLVING D ECIMALS

1.15 SOLVEMULTI-STE P E QUATIONS INVOLVING F RACTIONS

1.16 APPLICATIONS OF ONE -STE PEQUATIONS

1.17 SOLVEMULTI-STE P E QUATIONS INVOLVING R ATIONALN UMBERS

1.18 PROBLEM-SOLVING MODELS

1.19 COMPARISON OF P ROBLEM-SOLVING MODELS

1.20 DISTRIBUTIVEP ROPERTY FORM ULTI-S TEP EQUATIONS

1.21 SOLVING R EA L-WORLDP ROBLEMS USING MULTI-STE PEQUATIONS

Introduction

Here you will focus on solving equations. You will start with basic equations that can be solved in one step and

then move to more complicated equations that will require combining like terms and the distributive property. You

will also learn how to solve equations with variables on both sides and multi-step equations that might have rational

numbers in them.

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1.1. Simplifying Algebraic Expressions www.ck12.org

1.1 Simplifying Algebraic Expressions

Here youll learn to combine like terms within an expression.

Corey has a bowl of fruit that consists of 5 apples, 4 oranges, and 3 limes. Katelyn went to the farmers marketand picked up 2 apples, 5 limes, and an orange. How many apples, oranges, and limes do Corey and Katelyn have

combined?

Combining like terms is much like grouping together different fruits, like apples and oranges.

Watch This

MEDIAClick image to the left for more content.

James Sousa: Combining Like Terms

Guidance

You might have noticed from the previous concept, that sometimes variables and numbers can be repeated within an

expression. If the same variable is in an expression more than once, they can be combined by addition or subtraction.

This process is calledcombining like terms.

Example A

Simplify 5x 12 3x + 4.Solution: Reorganize the expression to group together thexs and the numbers. You can either place the like terms

next to each together or place parenthesis around the like terms.

5x

12

3x + 4

5x 3x 12 + 4or(5x 3x) + (12 + 4)2x 8

Notice that the Greatest Common Factor (GCF) for 2x and 8 is 2. Therefore, you can and use the Distributive

Property to pull out the GCF to get 2(x 4).

Example B

Simplify 6a 5b + 2a 10b + 7

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Solution: Here there are two different variables, a and b. Even though they are both variables, they aredifferent

variables and cannot be combined. Group together the like terms.

6a 5b + 2a 10b + 7(6a + 2a) + (5b 10b) + 7

(8a 15b + 7)There is only one number term, called the constant, so we leave it at the end. Also, in general, list the variables in

alphabetical order.

Example C

Simplifyw2 + 9 4w2 + 3w4 7w 11.Solution: Here we have one variable, but there are different powers (exponents). Like terms must have the same

exponent in order to combine them.

w2 + 9 4w2 + 3w4 7w 113w4 + (w2 4w2) 7w + (9 11)3w4 3w2 7w 2

When writing an expression with different powers, list the powers from greatest to least, like above.

Intro Problem RevisitedLets rewrite Coreys bowl of fruit as 5a + 4o + 3l, wherearepresents apples,orepresentsoranges, and l represents limes. Then Katelyns bowl of fruit can be represented as 2a + 5l+ o. Combining like

terms, we have:

(5a + 4o + 3l) + (2a + 5l + o)

(5a + 2a) + (4o + o) + (3l + 5l)

7a + 5o + 8l

Together they have 7 apples, 5 oranges, and 8 limes.

Guided Practice

Simplify the expressions below.

1. 6s 7t+ 12t 10s2. 7y2 9x2 +y2 14x + 3x2 4

Answers

1. Combine thess and thets.

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1.1. Simplifying Algebraic Expressions www.ck12.org

6s 7t+ 12t 10s(6s 10s) + (7t+ 12t) 4s + 5t

2. Group together the like terms.

7y2 9x2 +y2 14x + 3x2 4(9x2 + 3x2) + (7y2 +y2) 14x 46x2 + 8y2 14x 4

Notice in #1, we did not write(6s 10s) (7t+ 12t)in the second step. This would lead us to an incorrect answer.Whenever grouping together like terms and one is negative (or being subtracted), always change the operator to

addition and make the number negative.

In #2, we can also take out the Greatest Common Factor of -2 from each term using the Distributive Property. Thiswould reduce to 2(3x2 4y2 + 7x + 2). In this case, we take out a -2 so that the first term is positive.

Vocabulary

Constant

A number that is added or subtracted within an expression. In the expression 3x2 8x15, -15 is the constant.

Greatest Common Factor (GCF)

The largest number or variable that goes into a set of numbers.

Practice

Simplify the following expressions as much as possible. If the expression cannot be simplified, write cannot be

simplified.

1. 5b 15b + 8d+ 7d2. 6 11c + 5c 183. 3g2 7g2 + 9 + 124. 8u2 + 5u 3u2 9u + 145. 2a

5f

6. 7pp2 + 9p + q2 16 5q2 + 67. 20x 6 13x + 198. 8n 2 5n2 + 9n + 14

Find the GCF of the following expressions and use the Distributive Property to simplify each one.

9. 6a 1810. 9x2 1511. 14d+ 712. 3x 24y + 21

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Challenge We can also use the Distributive Property and GCF to pull out common variables from an expression.

Find the GCF and use the Distributive Property to simplify the following expressions.

13. 2b2 5b14. m3 6m2 + 11m15. 4y4 12y3 8y2

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1.2. Expressions and the Distributive Property www.ck12.org

1.2 Expressions and the Distributive Property

Here youll use the distributive property to write and evaluate equivalent numerical and algebraic expressions.

Have you ever been starving after physical activity?

One day after diving, Cameron and some of the kids that he has met at the resort decide to eat some hotdogs on the

beach. Cameron takes everyones order and heads to the hot dog stand. He figures that he will get a tray. Each of

the kids has given him a few dollars and Cameron thinks that he has enough money to get everything.

When he gets to the stand, he checks the prices. Cameron needs to buy 9 hotdogs. The hotdogs are $1.50 for a plain

dog plus 1.00 for cheese and sauce. Everyone wants cheese and sauce, so Cameron needs to buy nine hotdogs with

cheese and sauce.Given these numbers, how much will Cameron spend?

Cameron isnt sure. He takes a napkin and asks for a pen so that he can figure it all out. He has $25.00. Does he

have enough?

The Distributive Property will be very helpful to Cameron.

What is the Distributive Property? Well, this is the Concept that will teach you all about it. Pay attention and

at the end of this Concept you will help Cameron to get some lunch.

Guidance

The Distributive Property statesthat when a factor is multiplied by the sum of two numbers, we can multiply each

of the two numbers by that factor and then add them. You will see a term outside of the parentheses and then you

will know that we are dealing with the Distributive Property.

6(3 + 5)

5(x + 3)

You can use the Distributive Property to write equivalentexpressions. You know that equivalent means equal. Well,

we can write equivalent numerical and algebraic expressions using the Distributive Property.

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How do we write an equivalent numerical expression?

You write an equivalent numerical expression by writing the expression without the parentheses. To do this,

we multiply the term outside the parentheses with both terms inside the parentheses.

5(2 + 3)

Here we multiply five times 2 and times 3. Because this is over addition, the addition sign stays between the two

terms.

5(2 + 3) =5(2) + 5(3)

This is an equivalent numerical expression.

How do we write equivalent algebraic expressions?

Well, an algebraic expression is going to involve numbers, operations, variables and sometimes exponents too.

We simply take the term outside the parentheses and multiply it with both of the terms inside the parentheses.

4(x + 3)

Here we multiply four timesx and four times 3. Because this is over addition, the addition sign stays in the middle.

4(x + 3) =4(x) + 4(3)

This is an equivalent algebraic expression.

Take a few minutes to write down the steps in using the Distributive Property.

You can also use the Distributive Property to evaluate an expression. You might have caught yourself trying

to do that in the last Concept. Well, the first step is to write an equivalent expression as we did in the last

section and then we can simplify our work.

Lets start with a numerical expression. Because a numerical expression does not contain a variable, we will be

able to figure out an answer for the expression.

7(2 + 3)

First, we write an equivalent expression.

7(2 + 3) =7(2) + 7(3)Next, we multiply each part and then we add the products.

14 + 21

Our answer is 35.

This works the same way if there was subtraction involved.

3(3 2)First, we write an equivalent expression.

3(3) 3(2)

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Next, we evaluate the expression.

9 6Our answer is 3.

How does this work with an algebraic expression?

We can follow the same procedure, but keep in mind that an algebraic expression will have variables in it. Therefore,

we can simplify the expression, but not necessarily solve it.

2(x + 6)

First, we write an equivalent expression.

2(x) + 2(6)

Next, we simplify each part of the expression.

2x + 12

This is our answer.

5(y 2)First, we write an equivalent expression.

5(y) 5(2)Next, we simplify each part of the expression.

5y 10This is our answer.

Write an equivalent expression for each using the Distributive Property.

Example A

6(5 + 2)

Solution: 42

Example B

3(x

5)

Solution: 3x 15

Example C

8(9 +y)

Solution: 72 + 8y

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Here is the original problem once again.

One day after diving, Cameron and some of the kids that he has met at the resort decide to eat some hotdogs on the

beach. Cameron takes everyones order and heads to the hot dog stand. He figures that he will get a tray. Each of

the kids has given him a few dollars and Cameron thinks that he has enough money to get everything.

When he gets to the stand, he checks the prices. Cameron needs to buy 9 hotdogs. The hot dogs are $1.50 for a plain

dog plus 1.00 for cheese and sauce. Everyone wants cheese and sauce, so Cameron needs to buy nine hot dogs with

cheese and sauce.

Given these numbers, how much will Cameron spend?

Cameron isnt sure. He takes a napkin and asks for a pen so that he can figure it all out. He has $25.00. Does he

have enough?

The Distributive Property will be very helpful to Cameron.

Think about the Distributive Property. Lets use it to write an expression to help Cameron.

9(1.50 + 1.00)

Next, we can distribute the 9.

9(1.50) + 9(1.00)

9 1.50=13.509 1=913.50 + 9.00

=$22.50

Cameron will spend $22.50 on the hotdogs. Out of $25.00, he will have $2.50 change.

Vocabulary

The Distributive Property

this property states that when a term is outside of the parentheses, that you multiply the term outside the

parentheses with terms inside the parentheses. The property can be over addition or subtraction.

Equivalent

equal. Equivalent expressions are equal expressions.

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Guided Practice

Here is one for you to try on your own.

Liam has a rectangular backyard that is 20 yards long and 18 yards wide. He wants to use a part of his yard that is

20 yards by 8 yards for a vegetable garden. If he does this, what will be the area of the section of the yard that will

not be used as a garden?

Lets start by drawing a diagram of Liams backyard to help us understand this problem better.

Answer

One way we can find the area of the section that will not be used as a garden is by subtracting the area of the garden

from the total area of the yard.

Remember, to find the area of any rectangle, including a rectangular yard, multiply the length times the width.

(area of entire yard) (area of garden) = (area of section not used as a garden)

(20 18) (20 8) =?We can make the computation easier by using the distributive property. Since the factor 20 is multiplied by both of

the other numbers, we can rewrite the expression as the product of 20 and the difference of the other two numbers.

(20 18) (20 8) =20 (18 8) =20 10=200

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Video Review

MEDIA

Click image to the left for more content.

- This is a James Sousa video on the distributive property.

Practice

Directions: Use the Distributive Property to write an equivalent expression for each numerical expression.1. 6(3 + 4)

2. 5(4 + 1)

3. 12(3 + 5)

4. 6(7 + 8)

5. 2(4 + 5)

6. 3(5 2)7. 6(7 3)

8. 5(4 2)9. 7(5 1)10. 6(9 3)Directions: Use the Distributive Property to write an equivalent expression for each variable expression.

11. 5(x + 3)

12. 6(y 2)13. 5(x + 9)

14. 8(a + b)

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15. 7(x y)Directions: Use the Distributive Property to evaluate each numerical expression.

16. 6(3 + 4)

17. 5(4 + 1)

18. 12(3 + 5)

19. 6(7 + 8)20. 2(4 + 5)

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1.3 Translate Verbal Phrases into Variable Ex-pressions

Here youll learn to translate verbal phrases into variable expressions.

Have you ever had to figure out a math problem that was described in words? Look at this dilemma.Kerry and her brother sold smoothies and cookies at the PTA Family Fun Day. In two hours they sold the lemonade

for 50,000 VND per glass and the cookies for 20,000 VND a piece. When finished, Kelly realized that they had sold

fifty glasses of lemonade and twenty cookies. She said this to her brother.

"We sold fifty times 50,000 VND and twenty times 20,000 VND."

Jamie, Kerrys brother isnt sure how to write this expression. Pay attention and you will be able to help him at the

end of the Concept.

Guidance

Do you know how to take a verbal phrase and write it as a variable expression?

To accomplish this task, you will need to think about what different words mean. A verbal expression is a mathe-

matical statement that is expressed in words.

You will have to work as a detective to figure out what different words mean. Once you know what those words

mean, you will be able to write different variable expressions.

TABLE 1.1: Lets start by looking at some mathematical operations written as words.

Addition Subtraction Multiplication Division

Sum Difference Product Quotient

Plus Less than Times Split up

Increased by Take away Of Per

More Decreased by Multiplied by Ratio of

Total of Minus

added to Fewer Than

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1.3. Translate Verbal Phrases into Variable Expressions www.ck12.org

This list does not include ALL of the ways to write the operations, but it will give you a good place to start.

Take a few minutes and write these words down in your notebook.

Now we can look at the following chart which starts with a verbal phrase and writes it as a variable expression.

TABLE1.2:

Verbal Phrase Variable Expression

Three minus a number 3 xA number increased by seven n + 7The difference between an unknown quantity and

twenty-six

s 26

A number decreased by nine w

9

Ten times a number plus four 10f+ 4

Notice that words like a number and an unknown quantity let us know that we need to use a variable.

Example A

Write a variable expression that reads The product of a number and six plus four.

Solution: 6x + 4

Example B

Write a variable expression that reads Ninety divided by a number minus eight.

Solution: 90b 8

Example C

Write a variable expression that reads Two less than a number, multiplied by thirty-six.

Solution: 36(n 2)

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Now lets go back to the dilemma from the beginning of the Concept. Kelly explained the sales to her brother in this

way.

"We sold fifty times 50,000 VND and twenty times 20,000 VND."

First, use the information in the statement to write an expression.

50(50, 000) + 20(20, 000)

Notice that we have fifty times two dollars plus twenty times one dollar and fifty cents. This shows the number of

glasses of lemonade and cookies times each price.

Next, we can figure out how much money they made.

2, 900, 000

So Kerry and Jamie made 2.9 million VND in two hours.

Vocabulary

Variable Expression

a group of numbers, operations and variables without an equal sign.

Variable

a letter used to represent an unknown number.

Constant

a number in an expression that does not have a variable.

Verbal Expression

using language to write a mathematical expression instead of numbers, symbols and variables.

Guided Practice


Write a variable expression that reads Eighty-five divided by a number minus thirteen.

Solution

We could do this in several different ways. We could use a symbol, , to show division or we could use a fractionbar to show division.

Because you are moving toward Algebra, lets use a fraction bar.

The answer is 85a 13.

Video Review

MEDIA


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1.3. Translate Verbal Phrases into Variable Expressions www.ck12.org

Writing Basic Algebraic Expressions

Practice

Directions: Write a variable expression for each verbal expression.

1. The sum of a number and twelve.

2. The difference between a number and eight.

3. Three times a number

4. A number squared plus five

5. A number divided by two plus seven

6. Four times the quantity of a number plus six

7. A number times two divided by four

8. A number times six plus the same number times two

9. A number squared plus seven take a way four

10. A number divided by three plus twelve

More Practise

From IXL do Write variable expressions and Write Variables to Represent Diagrams (Make sure you go past

problems where you are combining like terms and adding or subtracting, this will make sense when you try a few).

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1.4 One-Step Equations and Inverse Opera-tions

Here youll learn how to isolate variables using inverse operations in order to solve equations in one step.

On the pretest you have all shown me that you can solve simple questions like x + 4= 16 or find the solution tothe equation 9x= 72 in one step. However, you also have shown me that you try to solve all problems throughinspection. Trying a number out then seeing if it works or just evaluating, yes 12 + 4=16 . This works well for theexamples I gave before but not so well for something like 2 (3x + 2)6x= (x + 4) + 16. So we will start from thebeginning and move towards more complicated equations!

Guidance

Its Easier than You Think

You have been solving equations since the beginning of this textbook, although you may not have recognized it. For

example, in a previous Concept, you determined the answer to the pizza problem below.$20.00was one-quarter of the money spent on pizza.

14 m=20.00 What divided by 4 equals 20.00?

The solution is 80. So, the amount of mo So weney spent on pizza was $80.00.

By working through this question mentally, you were applying mathematical rules and solving for the variablem.

Definition: To solvean equation means to write an equivalent equation that has the variable by itself on one side.

This is also known asisolating the variable.

In order to begin solving equations, you must understand three basic concepts of algebra: inverse operations,

equivalent equations, and the Addition Property of Equality.

Inverse Operations and Equivalent Equations

In another Concept, you learned how to simplify an expression using the Order of Operations:Parentheses,Exponents,

Multiplication and Division completed in order from left to right, and Addition and Subtraction (also completed from

left to right). Each of these operations has an inverse. Inverse operations undo each other when combined.

For example, the inverse of addition is subtraction. The inverse of an exponent is a root.

Definition: Equivalent equationsare two or more equations having the same solution.

The Addition Property of Equality

Just like Spanish, chemistry, or even music, mathematics has a set of rules you must follow in order to be successful.

These rules are called properties, theorems, or axioms. They have been proven or agreed upon years ago, so you can

apply them to many different situations.

For example, theAddition Property of Equalityallows you to apply the same operation to each side of an equation,

or what you do to one side of an equation you can do to the other.

The Addition Property of Equality

For all real numbersa, b,and c:

Ifa=b, thena + c=b + c.

Solving One-Step Equations Using Addition or Subtraction

Because subtraction can be considered adding a negative, the Addition Property of Equality also works if you need

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1.4. One-Step Equations and Inverse Operations www.ck12.org

to subtract the same value from each side of an equation.

Example A

Solve for y :

16=y 11.Solution: When asked to solve fory, your goal is to write an equivalent equation with the variable y isolated on oneside.

Write the original equation: 16=y 11.Apply the Addition Property of Equality: 16 + 11=y 11 + 11.Simplify by adding like terms: 27=y.

The solution isy=27.

Example B

Solve for z:

5=z + 12

Solution:

Apply the Addition Property of Equality:

5=z + 12

5 12=z + 12 125 12=z

7=zThe solution is 7=z.Equations that take one step to isolate the variable are called one-step equations. Such equations can also involve

multiplication or division.

Solving One-Step Equations Using Multiplication or Division

The Multiplication Property of Equality

For all real numbersa, b, andc:

If a=b,then a(c) =b(c).

Example C

Solve for k: 8k= 96.Solution: Because8k= 8 k, the inverse operation of multiplication is division. Therefore, we must cancelmultiplication by applying the Multiplication Property of Equality.

Write the original equation: 8k= 96.Apply the Multiplication Property of Equality: 8k8= 96 8.The solution isk=12.

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Video Review

MEDIA


Guided Practice

1. Determine the inverse of division.

2. Solve 18 x=1.5.Solutions:

1. To undo division by a number, you would multiply by the same number.

2. The variablex is being multiplied by one-eighth. Instead of dividing two fractions, we multiply by the reciprocal

of 18 , which is 8.

8

1

8x

=8(1.5)

x=12

Practice

Sample explanations for some of the practice exercises below are available by viewing the following video. Note

that there is not always a match between the number of the practice exercise in the video and the number of the

practice exercise listed in the following exercise set. However, the practice exercise is the same in both. CK-12 Ba

sic Algebra: One-Step Equations(12:30)

MEDIA


Solve for the given variable.

1. x + 11=72. x 1.1=3.23. 7x=214. 4x=15. 5x

12= 2

3

6. x +52 = 23

7. x 56 =

38

8. 0.01x=11

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1.4. One-Step Equations and Inverse Operations www.ck12.org

9. q 13= 1310. z + 1.1=3.000111. 21s=312. t+ 12=

13

13. 7f11 =

711

14. 34

= 12y

15. 6r= 38

16. 9b16= 38

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1.5 Solve Equations Involving Inverse Proper-ties of Addition and Multiplication

Here youll solve equations involving the inverse properties of addition and multiplication.

The marching band at Floyd Middle School is excellent and well respected throughout the community. Each year

the band continues to grow, and the students who participate make a huge commitment to practices, football games

and parades.

When the students gathered for their first rehearsal, Mrs. Kline the band director gathered them altogether for a few

announcements.

We will be marching in the big parade this year once again, she said smiling from ear to ear.

I am so glad, Keri said leaning over to Anica. I was hoping that we would.

We will also be adding four new students this year. We started with 140 students and we will add four. This means

that we need to redo our formation for the big finale. We need to reorganize the band into eight even rows. Lets

take a look at what this will look like. Please take out a piece of paper and a pencil, Mrs. Kline said turning to the

blackboard.

I can figure out the number of students in each row with an equation, Anica said smiling.

Yes, and dont forget to count Jake as the Drum Major in the lead, Keri added.

Do you know what this equation needs to look like? We have been given the sum of the students, a Drum

Major and we know that we need eight even rows. As Anica said, we will need an equation to figure out the

number of students in each row. This Concept will teach you all that you need to know about equations so

that you will know how to solve this in the end.

Guidance

What do you know about equations?

Anequationis a statement with an equal sign where the quantity on one side of the equals is the same as the

quantity on the other side of the equals.

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1.5. Solve Equations Involving Inverse Properties of Addition and Multiplication www.ck12.org

Here is a simple equation.

x + 11=15

Here we have an equation with a variable where x is the unknown quantity. To solve this, the natural thing to do is

to performaninverse operationor opposite operation. We subtract eleven from 15 which leaves us with 4. That is

the value of the variable.

Most of the time, you dont even think about performing an inverse operation, your mind naturally solves the

problem in this way.

When you have an equation with one variable, it is called a one-step equation. It only takes one operation or

one inverse operation to solve it. You have had a lot of practice solving one-step equations.

To solve a two-step equation, we will need to use more than one inverse operation.

Lets take a look at how to solve a two-step equation.

When we perform inverse operations to find the value of a variable, we work to get the variable alone on one side

of the equals. This is called isolating the variable. It is one strategy for solving equations. You can use isolating the

variable whether you are solving one-step or two-step equations.

Here is a two-step equation.

Solve for a: 3a + 12=45.

We can call each piece of the equation a term. There is a term with a variable and there is a term without a variable.

Notice that there are twotermson the left side of the equation, 3aand 12.

Our first step is to use inverse operations to get the term that includes a variable, 3a, by itself on one side of

the equal (=) sign. Because the three is connected to the variable, we perform the other inverse operation first.

We work with the number connected with the variable last.

In the equation, 12 is addedto 3a. So, we can use the inverse of additionsubtraction. We can subtract 12

from both sides of the equation.

Lets see what happens when we subtract 12 from both sides of the equation.

3a + 12=45

3a + 12 12=45 123a + 0=33

3a=33

Now, the term that includes a variable, 3a, is by itself on one side of the equation.

We can now use inverse operations to get the a by itself. Since 3a means 3 a, we can use the inverse ofmultiplicationdivision. We can divide both sides of the equation by 3. Lets see what happens when we

divide both sides of the equation by 3.

3a=33

3a

3 =

33

3

1a=11

a=11

The value ofa is 11.

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Lets review our steps to solving this two-step equation.

Take a few minutes to write these steps in your notebook.

Example A

4x + 5=29

Solution: x =6

Example B

3y + 7=43

Solution: y =12

Example C

6x + 8=71

Solution: x =9

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Now lets go back to the dilemma from the beginning of the Concept.

First, lets look at the given information.

There are 144 students in the band.

There is also one Drum Major.

We need to organize the students into eight even rows.

Here is our equation.8x=144

Now you may be wondering why we didnt include the Drum Major. Well, as Keri points out, the Drum Major

is in the lead. In this case, Jake is not included in the equation since he is not in the rows.

We have a one-step equation here. We can solve the equation now.

x=18students

There will be 18 students in each row.

Vocabulary

Equation

a mathematical statement with an equal sign where the quantity on one side of the equation is equal to the

quantity on the other side.

Variable

a letter used to represent an unknown quantity.

Algebraic Equation

An equation with at least one variable in it.

One-Step Equation

An algebraic equation with one operation in it.

Two-Step Equation

An algebraic equation with two operations in it.

Guided Practice


A gardener charges $20 for each gardening job plus $15 for each hour worked. He charged $80 for a gardening job

he did yesterday.

a. Write an algebraic equation to representh, the number of hours that the gardener worked on that $80 job.

b. Find the number of hours that the gardener worked on that $80 job.

Solution

Consider parta first.

Use a number, an operation sign, a variable, or an equal sign to represent each part of the problem. The gardener

earned $15 for each hour worked on that job, so you could multiply $15 by h, the number of hours worked, to find

how much money the gardener charged for his worktime.

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$20 f or each gardening job plus $15 f or each hour worked. . . charged$80 for one . . .job.

20 + 15h = 80

So, this equation, 20 + 15h=80, representsh, the number of hours the gardener worked on the $80 job.

Next, consider partb.

Solve the equation to find the number of hours the gardener worked on that job.

Since 20 is added to the term that includes a variable, 15h, we can use the inverse of additionsubtraction. We can

subtract 20 from both sides of the equation, like this:

20 + 15h=80

20 20 + 15h=80 200 + 15h=60

15h=60

Since 15 is multiplied by the variable, h, we can use the inverse of multiplicationdivision. We can divide both

sides by 15 to solve forh, like this:

15h=60

15h

15 =

60

15

1h=4

h=4

The gardener worked four hours on the job he did yesterday.

Video Review

MEDIA


Khan Academy Two-Step Equations

Practice

Directions: Solve the following two-step equations that have addition and multiplication in them.

1. 3x + 4=22

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2. 4y + 3=153. 6x + 5=354. 7x + 2=165. 9y + 8=806. 12x + 15=517. 14y + 2=308. 7y + 5=40

9. 2x + 4=4810. 6x + 3=3911. 8x + 2=1012. 8x + 7=9513. 9x + 9=9014. 3x + 5=5015. 7x + 12=61

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1.6 Solve Equations Involving Inverse Proper-ties of Addition and Division

Here youll solve equations involving the inverse properties of addition and division.

Jessica and Casey worked at a bakery during school vacation. One day Casey was asked to divide up many poundsof flour. She divided the amount she was given by three. Then she added four more pounds to one of these portions.

Jessica was given the largest portion. If Jessica received 8 pounds of flour, how many pounds of flour did Casey

begin with?

Do you know how to solve this problem? To figure it out, you will need to write and solve a two-step equation. Pay

attention to this Concept and you will know how to solve it by the end of the Concept.

Guidance

You are going to learn how to solve two-step equations. Lets begin.

To solve a two-step equation, we will need to use more than one inverse operation. Lets take a look at how to solve

a two-step equation now. When we perform inverse operations to find the value of a variable, we work to get the

variable alone on one side of the equals. This is called isolating the variable. It is one strategy for solving equations.

You can use isolating the variable whether you are solving one-step or two-step equations.

Solve for c: 5 + c4=15.

Notice that there are two terms on the left side of the equation, 5 and c4

.

Our first step should be to use inverse operations to get the term that includes a variable, c4

, by itself on one

side of the equal (=) sign.

In the equation, 5 isaddedto c4 . So, we can use the inverse of additionsubtraction. We can subtract 5 from

both sides of the equation.

5 +c

4= 15

5 5 +c4

= 15 5

0 +c

4= 10

c

4= 10

Now, the term that includes a variable, c4

, is by itself on one side of the equation.

We can now use inverse operations to get the c by itself. Since c4

means c 4, we can use the inverse ofdivisionmultiplication. We can multiply both sides of the equation by 4.

c

4=10

c

4 4=10 4

c

4 4

1=40

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1.6. Solve Equations Involving Inverse Properties of Addition and Division www.ck12.org

The number 4, or 41

, is the multiplicative inverse, orreciprocal, of 14

. You can find the multiplicative inverse of a

number by flipping its numerator and its denominator. So, the multiplicative inverse of 41

is 14

. When a number is

multiplied by its multiplicative inverse, the product is 1.

c4

41

=40

c

1

4 4

1

=40

c 1=40c=40

The work above shows how multiplying each side of the equation by 4 isolates the variable.

Because 4 is the multiplicative inverse, or reciprocal, of 14 , we could also have solved this problem by canceling outthe 4s like this:

c

4 4

1= 40

c

1= 40

c=40

The answer is thatc is equal to 40.


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Example A

x5+ 6=10

Solution: x =20

Example B

x9+ 12=28

Solution: x=144

Example C

x11

+ 12=18

Solution: x=66


Think about what we know. We know that Casey divided the pounds of flour by three, but we dont know how many

pounds she started with, so this our variable.

x

3

Next, we know that Casey added four pounds to one of the portions.

x3+ 4

Jessica ended up with 8 pounds.

x3+ 4=8

Now we can solve the equation. Start by subtracting four from both sides of the equation.

x3+ 4 4=8 4x3 = 4

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1.6. Solve Equations Involving Inverse Properties of Addition and Division www.ck12.org

Next we use the inverse of division, multiplication, and multiply three times four.

x=12

Casey started with twelve pounds of flour.

Vocabulary

Equationa mathematical statement with an equal sign where the quantity on one side of the equation is equal to the


Variable


Algebraic Equation


One-Step Equation


Two-Step Equation


Guided Practice


y19+ 6=10

Solution

First, we have to subtract 6 from each side of the equation.y19

=10 6y19

=4

Now we can multiply 19 times 4. This will give us the value ofy.

(19)(4) =76

y=76

This is our solution.

Video Review

MEDIA


Solving Two-Step Equations

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Practice

Directions: Solve the following two-step equations that have addition and division in them.

1. x3+ 4=8

2. x5+ 8=10

3. a6+ 7=13

4. a

9+ 4=305. b

8+ 6=15

6. c12

+ 9=187. x

7+ 7=218. x

11+ 5=129. x

12+ 9=1610. a

14+ 6=811. x

22+ 9=1212. y

2+ 14=18

13. x7+ 24=38

14. x8+ 15=30

15.

x

9+ 11=28

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1.7. Solve Equations Involving Inverse Properties of Subtraction and Multiplication www.ck12.org

1.7 Solve Equations Involving Inverse Proper-ties of Subtraction and Multiplication

Here youll solve equations involving inverse properties of subtraction and multiplication.

Have you ever looked at a homework problem and wondered how to solve it? Look at this situation that Henry faced.Henry looked at the first problem on his homework page.

14x 9=19Even though hed been paying attention in class, Henry had no idea how to solve this problem.

Do you know how to solve it? This is a two-step equation involving subtraction and multiplication. This Concept

will teach you the steps for solving equations like this one.

Guidance

You are going to learn how to solve two-step equations with subtraction and multiplication in them. Lets begin.To solve a two-step equation, we will need to use more than one inverse operation. Lets take a look at how to solve




Solve for x: 2x 9=17.Notice that there are two terms on the left side of the equation, 2xand 9. Our first step should be to use inverse

operations to get the term that includes a variable, 2x, by itself on one side of the equal (=) sign.

In the equation, 9 is subtractedfrom2x. So, we can use the inverse of subtractionaddition. We can subtract

9 from both sides of the equation.

2x 9=172x(9 + 9) =17 + 9

2x=26

Notice how we rewrote the problem above. Since we are adding a positive number, 9, to a number that is

being subtracted from2x, we can represent this as adding 9 to -9 as we did above: (-9 + 9).

The number 9 is theadditive inverse, or opposite, of -9.

We can now use inverse operations to get the x by itself. Since 2x means 2 x, we can use the inverse ofmultiplicationdivision. We can divide both sides of the equation by 2.

2x=26

2x

2 =

26

2

x=13

The value ofx is 13.

Lets review our steps for solving this two-step equation.

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Example A

9x 5=40Solution: x=5

Example B

9y

6=66

Solution: y=8

Example C

12a 4=44Solution: a=4


Here is the problem that Henry saw on his page.

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1.7. Solve Equations Involving Inverse Properties of Subtraction and Multiplication www.ck12.org

14x 9=19To solve this problem, we can first add nine to both sides of the equation.

14x 9 + 9=19 + 914x=28

Now Henry can solve this as a one-step equation by dividing both sides by 14.

x=2This is the answer to this problem.

Vocabulary

Equation



Variablea letter used to represent an unknown quantity.

Algebraic Equation


One-Step Equation


Two-Step Equation


Guided Practice


Eight times a number minus four is equal to ninety - two.

Write a two-step equation and solve for the missing variable.

Solution

First, walk through the words to write the equation.8x 4=92Now solve the for the variable. First, add four to both sides of the equation.

8x 4 + 4=92 + 48x=96

Now divide both sides by 8.

x=12

This is our answer.

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Video Review

MEDIA


[www.youtube.com/watch?v=9ITsXICV2u0 Solving Two-Step Equations]

Practice

Directions: Solve each two-step equation that has multiplication and subtraction in it.

1. 4x 3=132. 5y 8=223. 7x 11=314. 8y 15=255. 9x 12=426. 12y 9=997. 2y 3=238. 3x 8=199. 5y 2=28

10. 7x 11=3811. 5y 9=5112. 6a 12=3013. 9x 14=1314. 12x 23=4915. 13y 3=2316. 18x 12=42

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1.8. Solve Equations Involving Inverse Properties of Subtraction and Division www.ck12.org

1.8 Solve Equations Involving Inverse Proper-ties of Subtraction and Division

Here youll solve equations involving inverse properties of subtraction and division.

Did you ever solve a dilemma about wrapping paper? Take a look at this one.

Brandon and Felicia sold rolls of wrapping paper for a school fundraiser. Brandon sold 3 less than half the number

of rolls that Felicia sold. Brandon sold a total of 9 rolls of wrapping paper.

Write an algebraic equation to represent f, the number of rolls of wrapping paper that Felicia sold. Then, find the

number of rolls of wrapping paper that Felicia sold.

Do you know how to solve this dilemma? Notice that there will be two parts to your answer. Pay attention to this

Concept and you will know how to figure out this problem.

Guidance

You are going to learn how to solve two-step equations with subtraction and division. Lets begin.

To solve a two-step equation, we will need to use more than one inverse operation. Lets take a look at how to solve




Solve for z: z67=3.

Notice that there are two terms on the left side of the equation, z

6

and 7. Our first step should be to use inverse

operations to get the term that includes a variable, z6

, by itself on one side of the equal (=) sign.

In the equation, 7 is subtractedfrom z6

. So, we can use the inverse of subtractionaddition. We can add 7 to

both sides of the equation, like this:

z

6 7=3

z

6 7 + 7=3 + 7

z

6+ (7 + 7) =10z

6+ 0=10

z

6= 10

Now, the term that includes a variable, z6


We can now use inverse operations to get the z by itself. Since z6

means z 6, we can use the inverse of divi-sionmultiplication. We can multiply both sides of the equation by 6, like this:

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z

6=10

z

6 6=10 6

z

6 6

1=60

z1

=60

z=60

The value ofz is 60.



Example A

x38=9

Solution: x=51

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Example B

y72=13

Solution: y=105

Example C

a72=12

Solution: a=98

Now lets go back to the dilemma at the beginning of the Concept.

Consider partafirst.

Use a number, an operation sign, a variable, or an equal sign to represent each part of that problem. Since Brandon

sold 9 rolls of wrapping paper, represent the number of rolls Brandon sold as 9. Use the key words from the chart

to help you translate the rest of the problem into an equation. For example, you can translate half the number of

rolls...Felicia sold as f2

.

Brandon sold3 less than hal f the number. . . Felicia sold.

9 =

f

2 3

So, this equation, 9= f23, represents f, the number of rolls of wrapping paper that Felicia sold.


Solve the equation for f to find the number of rolls that Felicia sold.

Our first step should be to use inverse operations to get the term that includes a variable, f2

, by itself on one side of

the equal (=) sign. In the equation, 3 is subtractedfrom f2 . So, we can use the inverse of subtraction and add 3 to

both sides of the equation, like this:

9= f

23

9 + 3= f

23 + 3

12= f2

+ (3 + 3)

12= f

2+ 0

12= f

2

Now, the term that includes a variable, f2


We can now use inverse operations to get the f by itself. Since f2

means f 2, we can use the inverse of divisionand multiply both sides of the equation by 2, like this:

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12= f

2

12 2= f2

2

24= f

2 2

1

24= f1

24= f

The value of fis 24, so Felicia sold 24 rolls of wrapping paper for the fundraiser.

Vocabulary

Equation



Variable


Algebraic Equation


One-Step Equation


Two-Step EquationAn algebraic equation with two operations in it.

Guided Practice


x69=8

Solution

To solve this problem, first we have to add nine to both sides of the equation.

x

6 9 + 9=8 + 9x6

= 17

Next, multiply 6 times 17.

x=102

This is the answer.

Video Review

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MEDIA


Solving Two-Step Equations

Practice

Directions: Solve each two-step equation that has division and subtraction in it.

1. x54=8

2. y63=8

3. x77=10

4. x8 4=12

5. y

7 5=11

6. x410=12

7. y48=2

8. x312=9

9. a53=11

10. b41=15

11. x28=4

12. a7 4=9

13. b47=3

14. x81=12

15. y68=5

16. x2

15=12

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1.9 Solve Equations Involving Combining LikeTerms

Here youll learn to solve equations involving combining like terms.

Have you ever had a long rehearsal for something? Take a look at what is happening at band practice.

Wow, that was quite a day rehearsal, Jake said as he put his things away in the band room.

I agree. Im beat, Anica said.

We rehearsed for longer today than we did yesterday, Jake said.

Yes, 45 minutes longer. So we rehearsed for a total of five hours counting yesterday and today, Anica said.

Wait a minute you left me in the dust. How many minutes did we rehearse yesterday, and how many did we rehearse

today? Jake asked sitting down in a band chair.

Okay, let me show you. You need an equation, Anica said taking out a piece of paper and a pencil.

Do you know how Anica can figure this out? You will once you know how to work with multi-step equations.

Pay attention to this Concept and you will be able to figure out how long the rehearsals were at the end of it.

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1.9. Solve Equations Involving Combining Like Terms www.ck12.org

Guidance

Consider this simple problem.

Suppose you bought 3 yellow peaches, 5 white peaches and 2 red apples at a fruit stand.

You could also say that you bought 8 peaches and 2 apples because:

3 peaches + 5 peaches + 2apples=8 peaches + 2apples.

However, you couldnt say that bought 10 peaches. When you determine how many peaches you bought, you can

add 3 yellow peaches to 5 white peaches, but you cannot add the 2 red apples to that total. That is because apples

and peaches aredifferentkinds of fruit.

When you add or subtract the terms in an expression, you can only combine like terms.

Consider this expression:

3p + 5p + 2a

This expression represents the problem above. The variable p stands for peaches. The variablea stands for apples.

Just as you can combine the yellow peaches with the white peaches because they are both peaches, you can combine

3pand 5pbecause they are like terms. Each of those terms includes the same variable, p. However, you could not

combine 5pwith 2a, because they are not like terms. Each of those terms has a different variable.

Like termsare terms that contain the same variable, and these terms can be combined.

This shows how you could simplify the expression above by combining like terms:

3p + 5p + 2a=8p + 2a.

Lets take a look at how we can apply what we know about combining like terms to solving algebraic equa-tions.

Solve for r: 5r r9=15.First, combine the like terms5rand ron the left side of the equation. It may help to remember that r=1r.

5r r9=15(5r 1r) 9=15

4r9=15

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Notice that 9 cannot be combined with 4rbecause they arenotlike terms.

Now that we have combined like terms, we can solve the equation as we would solve any two-step equation.

Our next step is to isolate the term with the variable, 4r, on one side of the equation. Since 9 issubtractedfrom

4r, we should add 9 to both sides of the equation to isolate that term.

4r 9=154r 9 + 9=15 + 9

4r+ (9 + 9) =244r+ 0=24

4r=24

Since 4rmeans 4r, we should divide each side of the equation by 4 to get therby itself on one side of the equation.

4r=24

4r

4 =

24

4

1r=6

r=6

The value ofris 6.

Solve for n: 6n + 3 + 8n + 2=33.

First, combine the like terms on the left side of the equation. The terms 6nand 8nare like terms since each

has the same variable,n. The numbers 3 and 2 are also like terms, so they can be combined as well.

Use thecommutative property of additionto help you reorder the terms being added. This property states that termscan be added in any order. Then use theassociative property of addition to group the terms so like terms are being

added. The associative property of addition states that the grouping of terms being added does not matter.

6n + 3 + 8n + 2=33

6n + (3 + 8n) + 2=33

6n + (8n + 3) + 2=33

(6n + 8n) + (3 + 2) =33

Now, that the like terms are grouped together with parentheses, combine them.

(6n + 8n) + (3 + 2) =33

14n + 5=33

Now, we can solve as we would solve any two-step equation.

The next step is to isolate the term with the variable, 14n, on one side of the equation. Since 5 isaddedto 14n,

we should subtract 5 from both sides of the equation to do this.

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14n + 5=33

14n + 5 5=33 514n + 0=28

14n=28

Since14nmeans14 n, we should divide each side of the equation by 14 to get the n by itself on one side ofthe equation.

14=28

14n

14 =

28

14

1n=2

n=2

The value ofn is 2.

Example A

3p + 5p + 2=18

Solution: p=2

Example B

13x + 6x + 14a 9aSolution: 19x + 5a

Example C

3x + 5x + 9x 7=44Solution: x=3


First, we need to name the variable. We are looking to figure out times, so we can use tas our variable.

t= time

Next, we can write an equation. We know that there are two times.

t+ t

But also, one day was 45 minutes longer.

t+ t+ 45

The total sum of time is 5 hours. We need to make sure both our units are the same, so we convert 5 hours

into minutes and write 300 minutes.

t+ t+ 45=300

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Here is our equation.

Next, we solve it for the two times.

t+ t+ 45=300

2t= 300 452t= 255

t= 127.5minutes

This is the time that the band rehearsed yesterday. They rehearsed 45 more minutes today. We add 45 to the

total time from yesterday.

127.5 + 45=172.5minutes

Vocabulary

Like Terms

terms that include a common variable.

Commutative Property of Addition

states that the order that you add different numbers does not change the sum.

Guided Practice


Yesterday, Tanya biked 3 more miles than she biked today. She biked a total of 13 miles on both days.

a. Lettstand for the number of miles Tanya biked today. Write an algebraic equation to represent the number of

miles Tanya biked on both days.

b. Find the number of miles Tanya biked today.

c. Find the number of miles Tanya biked yesterday.

Solution

Consider partafirst.

You know thattrepresents the number of miles Tanya biked today. Use that variable to write an expression for the

number of miles Tanya biked yesterday.

Yesterday, Tanya biked3 more . . .than she biked t oday.

3 + t

So, you know that Tanya bikedtmiles today and 3 + tmiles yesterday. You also know that she biked a totalof 13miles on both days. Use this information to write an addition equation for this problem.

(miles biked today) + (miles biked yesterday) = (total miles biked)

t + 3 + t = 13

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So, this problem can be represented by the equation,t+ 3 + t=13.


The variabletrepresents the number of miles Tanya biked today. So, solve the equation for t.

First, use the commutative property of addition to rearrange the terms being added so it is easier to see how to add

the like terms.

t+ (3 + t) =13

t+ (t+ 3) =13

Now, add the like terms on the left side of the equation.

t+ t+ 3=13

(t+ t) + 3=13

2t+ 3=13

Solve the equation fortas you would solve any two-step equation. Subtract 3 from both sides of the equation.

2t+ 3=13

2t+ 3 3=13 32t+ 0=10

2t=10

Then, divide both sides of the equation by 2.

2t= 10

2t

2 =

10

2

1t= 5

t= 5

The value oftis 5, so Tanya biked 5 miles today.

Consider partc next.

In parta, you determined that Tanya biked 3 + tmiles yesterday. Sincet= 5, substitute 5 fortin the expression tofind how many miles she biked yesterday.

3 + t= 3 + 5=8

Tanya biked 8 miles yesterday.

Video Review

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MEDIA


Solving Multi-Step Equations

Practice

Directions: Practice combining like terms as you simplify each expression.

1. 8x + 3x + 22. 5y 3y + 83. 6x + 9x +x 44. 9x + 4x 8 + 2x5. 2y

10y + 16

6. 3x + 4x + 5 6 + 2x

Directions: Combine like terms and solve each equation.

7. 8x + 3x + 2=248. 5y + 2y + 6=489. 4x 6x + 3=13

10. 7y 10y + 6=911. 5x + 8x + 4=3012. 9a + 3a 4=4413. 7a + 4a + 6=83

14. 12x 14x + 3=1915. 10y 16y + 5=35

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1.10. Solve Equations with the Distributive Property www.ck12.org

1.10 Solve Equations with the Distributive Prop-erty

Here youll learn to solve equations with the distributive property.

Have you ever needed the distributive property to solve a problem? Well, Trevor has a dilemma. He is havingdifficulty figuring out this problem.

7(x + 2) =28

Do you know how to solve this equation? To figure it out, you will have to apply the distributive property. Take a

look at this Concept and you will know how to solve this equation by the end of it.

Guidance

You already know that some number properties can help you solve equations.

Thedistributive propertymay also help you solve some equations.

This property states that when a factor is multiplied by the sum of two numbers, we can multiply each of the two

numbers by that factor and then add them.

7 (4 + k) = (7 4) + (7 k) =28 + 7k2(a + 3) = (2 a) + (2 3) =2a + 6Multiplication can also be distributed over subtraction.

Here are two situations that show the distributive property.

7 (4 k) = (7 4) (7 k) =28 7k2(a

3) = (2

a)

(2

3) =2a

6

Lets see how the distributive property can help us solve some multi-step equations.

Solve for k:5(3 + k) =45

Apply the distributive property to the left side of the equation. Multiply each of the two numbers inside the

parentheses by 5 and then add those products.

5(3 + k) =45

(5 3) + (5 k) =4515 + 5k=45

Now, solve as you would solve any two-step equation. To get 5kby itself on one side of the equation, subtract 15

from both sides.

15 + 5k=45

15 15 + 5k=45 150 + 5k=30

5k=30

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To getkby itself on one side of the equation, divide both sides by 5.

5k=30

5k

5 =

30

5

1k=6

k=6

The value ofkis 6.

Lets look at another one.

2(y 9) =40Now we can distribute the two by multiplying it by both of the terms inside the parentheses. Notice that the second

term has a subtraction sign in front of it. Remember to include that sign when we multiply.

2y 18=40Next, we solve this for yas we would with any two step equation.

2y 18=402y 18 + 18=40 + 18

2y=58

y=29

The value ofy is 29.

Example A

6(x + 4) =42

Solution: x=3

Example B

4(y 8) =16Solution: y=12

Example C

12(x 2) =48Solution: x=6

Now lets go back to the dilemma at the beginning of the Concept.

7(x + 2) =28

This is the equation that needs to be solved.

First, we have to simplify the left side of the equation by getting rid of the parentheses. We do this by multiply both

of the terms inside the parentheses by 7.

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1.10. Solve Equations with the Distributive Property www.ck12.org

7x + 14=28

Next, we solve this two-step equation. Subtract 14 from both sides of the equation.

7x + 14 14=28 147x=14

Now we can solve the one-step equation by dividing both sides of the equation by 7.

x=2This is our final answer.

Vocabulary

Distributive Property

states that you can multiply a term outside of a set of parentheses with the terms inside the parentheses to

simplify the set of parentheses.

Guided Practice


Solve forx: 3(3 x) =12Solution

Apply the distributive property to the left side of the equation. Multiply each of the two numbers inside the

parentheses by 3 and then subtract those products. It may help you to remember thatx=1x.

3(3 x) =12

3(3 1x) =12(3 3) (3 1x) =12

9 3x=12

Now, solve as you would solve any two-step equation. We need to first get the term that includes a variable, 3x, by

itself on one side of the equation. In the equation, 3xis subtracted from 9. Subtracting 9 3xis the same as adding9 + (3x). Rewrite the left side of the equation to show that 9 is being added to 3x, and then subtract 9 from bothsides.

9

3x=12

9 + (3x) =129 9 + (3x) =12 9

0 + (3x) =33x=3

To get x by itself on one side of the equation, divide both sides by -3. You will need to use what you know about

dividing integers to help you. For example, you know that when you divide two negative integers, the quotient will

be positive. Since you know that 3 (3) =1, you also know that 3x (3) =1x. To review how to computewith integers, look back at Lessons 2.5 and 2.6.

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3x=33x3 =

3

31x= 1x=

1

The value ofx is -1.

Video Review

MEDIA


Khan Academy The Distributive Property

Practice

Directions: Use the distributive property to solve each equation.

1. 2(x + 3) =10

2. 5(x + 4) =253. 9(x 3) =274. 7(x + 5) =705. 5(x 6) =456. 8(y 4) =407. 7(x + 3) = 78. 8(x 2) =89. 9(y + 1) =90

10.3(y + 4) =2411.2(y 4) =1612.4(x 1) =813. 9(y

4) =36

14. 7(y 3) =2115.9(y 2) =27

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1.11. Solve Equations with the Distributive Property and Combining Like Terms www.ck12.org

1.11 Solve Equations with the Distributive Prop-erty and Combining Like Terms

Here youll solve equations with the distributive property and combining like terms.

Do you like candy? Take a look at this yummy dilemma.Eight children were given some candy. Then six different children were given the same unknown amount of candy.

Next, two children were that same unknown amount of candy plus three additional pieces of candy. The total number

of pieces of candy given out was thirty -eight.

If this is the case, what is the unknown amount of candy?

Do you know how to solve this problem? Write an equation and then solve it for the unknown amount of candy. Pay

attention and you will see this dilemma at the end of the Concept.

Guidance

To solve some multi-step equations you will need to use the distributive property and combine like terms. When thishappens, you will see that there is more than one term with the same variable or there is more than one number in

the equation. You always want to combine everything that you can before moving on to solving the equation.

Lets apply this to the following situation.

Solve for m: 6(1 + 2m) 3m=24Apply the distributive property to the left side of the equation. Multiply each of the two numbers inside the

parentheses by 6 and then add those products.

6(1 + 2m)

3m=24

(6 1) + (6 2m) 3m=246 + 12m3m=24

Next, subtract the like terms12mand 3mon the left side of the equation.

6 + 12m3m=246 + (12m 3m) =24

6 + 9m=24

Finally, solve as you would solve any two-step equation. Subtract 6 from both sides of the equation.

6 + 9m=24

6 6 + 9m=24 60 + 9m=18

9m=18

Now, divide both sides of the equation by 9.

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9m=18

9m

9 =

18

9

1m=2

m=2

The value ofm is 2.

Here is another one.

Solve for b: 4(2 + 3b) + 5b=13Apply the distributive property to th

math 8 supplemental text

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