2.3 i solving equations using subtraction and addition properties … · 2.3.2 model the...

Chapter 2 | lhe Language of Algebra

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the Subtraction and

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2.3 I Solving Equations Using

Addition Properties of Equality

Be Prepared!

By the end of this section, you will be able to:

2.3.1 Determine whether a number is a solution of an equation

2.3.2 Model the Subtraction Property of Equality

2.3.3 Solve equations using the Subtraction Property of Equality

2.3.4 Solve equations using the Addition Property of Equality

2.3.5 Translate word phrases to algebraic equations

2,3,6 Translate to an equation and solve

Learning Objectives

Before you get started, take this readiness quiz.

1-. Evaluate¡* Swhen x = ll.If you missed this problem, review Example 2.13

2. Evaluate5x- 3 whenx = 9.

If you missed this problem, review Example 2.L4

3. Translate into algebra: the difference of x and 8.

If you missed this problem, review Example 2.24

Ðetermine Whether a Number is a Solution of an EquationWhen some people hear the word algebra, they think of solving equations. The applications of solving equations are

limitless and extend to all careers and fields. In this section, we will begin solving equations. We will start by solving basicequations, and then as we proceed through the course we will build up our skills to cover many different forms of equations.

Solving an equation is like discovering the answer to a puzzle. An algebraic equation states that two algebraic expressions

are equal. To solve an equation is to determine the values of the variable that make the equation a true statement. Anynumber that makes the equation true is called a solution of the equation. It is the answer to the puzzle!

Solution of an Equation

A solution to an equation is a value of a variable that makes a true statement when substituted into the equation.

The process of finding the solution to an equation is called solving the equation.

To find the solution to an equation means to find the value of the variable that makes the equation true. Can you recognize

thesolution of x*2=7? lfyousaid 5, you'reright! Wesay 5 isasolutiontotheequation xI2= 7 becausewhen

we substitute 5 for x the resulting statement is true.

x*2=7s+zJt

I ='7 /

Since 5*2=7 isatruestatement,weknowthat 5 isindeedasolutiontotheequation.

I48

***Ç$"[g"d. ffig:"a rtThe symbol 3 asks whether the left side of the equation is

equal sign (=) or not-equal sign (#).

w Determine wh.th"r a number is a solution to an equation.

Step 1. Substitute the number for the variable in the equation.

Step 2. Simplify the expressions on both sides of the equation.

Step 3. Determine whether the resulting equalion is true.

(a) If it is true, the number is a solution.

(b) If it is not true, the number is not a solution.

Determine whether "r = 5 is a solution of 6x - l7 = 16.

Solution

Subst¡tute 5 for x.

Multiply.

Subtract.

6x-17=1626.5-17å16')30-17=:16

13+16

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So r = 5 is not a solution to the equation 6x - l7 = 16.

2.55 Isx = 3a solution of 4x-7 = 16?

2.56 ls x = 2a solution of 6; - 2 = l0?

Determine whethery = 2is a solution of 6y- 4 = 5y -2.

Solution

Example 2.28

Example 2.29

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Chapter 2 | The Language of Algebra r49

Design D YAftHere, the variable appears on both sides of the equation. We must substitute 2 for each y.

6y-4=5y-2

Substitute 2lor y. 6(2) - 4? sQ) - z

Mulriply. L2*43::d*ZSubtract. B=B{

Since y=2 resultsinatrueequation,weknowthat 2 isasolutiontotheequation 6y-4=5y-2.

2.57 Isy = 3 a solution of 9y - 2:8y+ l?H

B 2.58 Isy - 4a solution of 5y- 3 =3y*5?

Model the Subtract¡on Property of EqualityWe will use a model to help you understand how the process of solving an equation is like solving a puzzle. An enveloperepresents the variable - since its contents are unknown - and each counter represents one.

Suppose a desk has an imaginary line dividing it in half. We place three counters and an envelope on the left side of desk,and eight counters on the right side of the desk as in Figure 2.3. Both sides of the desk have the same number of counters,but some counters are hidden in the envelope. Can you tell how many counters are in the envelope?

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Figure 2,3

What steps are you taking in your mind to figure out how many counters are in the envelope? Perhaps you are thinking "Ineed to remove the 3 counters from the left side to get the envelope by itself. Those 3 counters on the left match with 3

on the right, so I can take them away from both sides, That leaves five counters on the right, so there must be 5 counters

in the envelope." Figure 2.4 shows this process,

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What algebraic equation is modeled by this situation? Each side of the

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.ffi.'ffi*$ "#'å,"n and the cenrer,inetakes the place of the equal sign. We will call the contents of the envelope x, so the number of counters on the left side

of thedeskis;*3. Ontherightsideof thedeskare 8 counters.Wearetoldthat r+3 isequalto 8 soourequationis

"x*3=8.

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Figure 2.5

x*3=8Let's write algebraically the steps we took to discover how many counters were in the envelope.

x*3=BFirst,wetookawaythreefrom each side. x + 3 * 3 = B - 3

Then we were left with five. x = 5

Now let's check our solution. We substitute 5 for -r in the original equation and see if we get a true statement.

x+3=B

s+338B=8r'

Our solution is correct. Five counters in the envelope plus three more equals eight.

MM Doing the ManipuÌative Mathematics activity, "Subtraction Property of Equality" will help you develop a betterunderstanding of how to solve equations by using the Subtraction Property of Equality.

Write an equation modeled by the envelopes and counters, and then solve the equation:

Solution

Example 2.30

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Chapter 2 | The Language of Alqebra

On the left, write x for the contents of the

envelope, and add the 4 counters, so we

havex* 4.

On the right, there are 5 counters.

The two sides are equal.

Solve the equation by subtracting 4

counters from each side.

151

Design D raftx*4

5

x*4=5

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We can see that there is one counter in the envelope. This can be shown algebraically as

x*4=5x14*4=5-4

x=L

Substitute I for ¡ in the equation to check.

x*4=521+4.:5

5=5{

Since .x = I makes the statement true, we know that I is indeed a solution.

2.59 Write the equation modeled by the envelopes and counters, and then solve the equation:

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r52 ê Chapter2lThe -il)^fffiflåH 2.60 Write the equation modeled by the envelopes and

Solve Equations Using the Subtraction Property of EqualityOur puzzle has given us an idea of what we need to do to solve an equation. The goal is to isolate the variable by itself onone side of the equations. In the previous examples, we used the Subtraction Property of Equality, which states that whenwe subtract the same quantity from both sides of an equation, we still have equality.

Subtraction Property of Equality

For any numbers a, b, and c, if.

a=bthen

a-c=b-c

Think about twin brothers Andy and Bobby. They are 17 years old. How old was Andy 3 years ago? He was 3 years less

than 17, sohisagewas 17-3; or 14. WhataboutBobby'sage 3 yearsago?Of course,hewas 14 also.Theirages

are equal now, and subtracting the same quantity from both of them resulted in equal ages 3 years ago.

a=ba-3=b-3

w Solve an equation using the Subtraction Property of Equality.

Step 1. Use the Subtraction Property of Equality to isolate the variable


Step 3. Check the solution.

Solve:x*8=17

Solution

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Example 2.33-

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Chapter 2 | The Language of Algebra

Subtract 74trom both sides

Simplify.

Substitute 26 for y to check.

1gg=y*742tæ=26+74

100 = 100 r'

Design Df'aftWe will use the Subtraction Property of Equality to isolate ¡.

x*B=L7SubtractBfrombothsides x+ I - B = L7 - B

Simplity. x = Ix*B=L79*B=1"7

L7 =17 I

Since x:9 makes x*8= 17 atruestatement,weknow 9 isthesolutiontotheequation.

2.61 Solve:

x*6:19

2.62 Solve:

x*9 = 14

Solve:100=y*74.

SolutionTo solve an equation, we must always isolate the variable-it doesn't matter which side it is on. To isolate y,

we will subtract 74 ftom both sides.

1çg=y*74l_00_74:y+74_74

26=y

Example 2.32

Since y = 26 makes 100 = )l + 74 a true statement, we have found the solution to this equation.

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2.63 Solve:

QJ=y+67

2.64 Solve

Çl=y+45

Solve Equations Using the Addition Property of EqualityIn all the equations we have solved so far, a number was added to the variable on one side of the equation. We usedsubtraction to "undo" the addition in order to isolate the variable. But suppose we have an equation with a number subtractedfrom the variable, such as x - 5 = 8. We want to isolate the variable, so to "undo" the subtraction we will add the numberto both sides. We use the Addition Property of Equality, which says we can add the same number to both sides of theequation without changing the equality. Notice how it minors the Subtraction Property of Equality.

Addition Propertv of Equalitv

For any numbers a, b, and, c, iÎ

a=bthen

ø+c=b+c

Remember the l.7-year-old twins, Andy and Bobby? In ten years, Andy's age will still equat Bobby's age. They will both

be 27.

a=ba+10=b+lO

We can add the same number to both sides and still keep the equality

w Solve an equation using the Addition Property of Equality.

Step 1. Use the Addition Property of Equality to isolate the variable.


Step 3. Check the solution.

Solve: x-5=8.

Solution

Example 2.33

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We will use the Addition Property of Equality to isolate the variable

x-5=BAdd5to bothsides x- 5 + 5 = B * 5

Simplify. x = 13

Nowwecancheck. Letx= 13. x- 5 = B

213-5å8' 8=8{

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2.65 Solve:

x-9=13

2.66 Solve:

y-l=3

Solve:27:a-16.

Solution

We will add 16 to each side to isolate the variable.

Add 16 to each side.

Simplify.

Nowwe can check. Leta = 43.

The solution to 27 = a - 76 ís a = 43

2.67 Solve:

19=a-18

2.68 Solve

27 =n-14

27=a-L627+16=a-1"6*16

43=a27*a-LE

e27 = 43- L6

27 =27 {

Example 2.34

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¡¡ Chapter 2 I The Lan&!ægof Algebra fl g

9ffiffi H__,Jr.ff*T".ilTranslate Word Phrases to AlgebraiRemember, an equation has an equal sign between nvo algebraic expressions. So if we have a sentence that tells us that twophrases are equal, we can translate it into an equation. We look for clue words that mean equols. Some words that translateto the equal sign are:

. is equal to

. is the same as

.is

. gives

. was

. will be

It may be helpful to put a box around the equalsword(s) in the sentence to help you focus separately on each phrase. Thentranslate each phrase into an expression, and write them on each side of the equal sign.

We will practice translating word sentences into algebraic equations. Some of the sentences will be basic number facts withno variables to solve for. Some sentences will translate into equations with variables. The focus right now is just to translatethe words into algebra.

Translate the sentence into an algebraic equation: The sum of 6 and 9 is equal to 15

SolutionThe word is tells us the equal sign goes between 9 and 15.

Locate the "equals" word(s).

Write the : sign

Translate the words to the left of theequars word into an algebraic expression.

Translate the words to the right of the equalsword into an algebraic expression.

The sum of 6 and I ß 15I

The sum of 6 and I = 15

6+9:

6+9=15

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2.69 Translate the sentence into an algebraic equation:

The sum of 7 and 6 gives 13.


The sum of 8 and 6 is 14.

Translate the sentence into an algebraic equation: The product of 8 and 7 is 56.

Solution

Example 2.35

Example 2.36

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chapter 2 I The Language of Algebra

The location of the word rb tells us that the equalsign goes between 7 and 56.


Write the = sign.

Translate the words to the left of theeqoals word into an algebraic expression.

Translate the words to the r¡ght of the equalsword into an algebraic expression.

ryryËfrüre ry e# dÉ*$

r(#tre

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The product of B and 7 ß56.I

The product of B and 7 : 56.

8.7 *

r57

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Recognize the key words:lwice; difference of .... and ...

Translate.

8.7=56

Twice the dilference of x and 3lgìtesl18.

Iw'ce means two limes.


The product of 6 and 9 is 54.


The product of 21 and 3 gives 63.

Translate the sentence into an algebraic equation: Twice the difference of x and 3 gives l8

Solution

Twice the dilference of x and 3 ls¡Veslra.

2 (x-3) 1B

H 2.73 Translate the given sentence into an algebraic equation:

Twice the difference of x and 5 gives 30.

H 2.74 Translate the given sentence into an algebraic equation:

Twice the difference of y and 4 gives 16.

Translate to an Equation and SolveNow let's practice translating sentences into algebraic equations and then solving them. We will solve the equations byusing the Subtraction and Addition Properties of Equality.

Example 2.37

Example 2.38

158 I Chapter2lThe Algebra

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Translate and solve: Three more than ¡ is equalto 47

Solution

Translate.

Subtract 3 from both sides of the equation.

Simplify.

we can check. Letx = 44.

So x = 44 is the solution.

2.75 Translate and solve:

Seven more than x is equal to 37


Eleven more than y is equal to 28

Translateandsolve:Thedifferenceof y and 14 is 18.

Solution

Three more than x is equal to 47

x*3=47x+3-3=47-J

x-44x*3=47

244+3¿47

47=47r'

Translate.

Add 14 to both sides.

Simplify.

We can check. Lety = 32.

So y = 32 is the solution.

The difference of y and 1-4 is 1-8.

y-t4=L8y-t4*14=L8+\4

Y=32y-L4=I8

82 - t43 L8

LB=lB/

Example 2.39

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The difference of z and 17 is equal to 37

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Thedifferenceof x and 19 isequalto 45

We encourage you to go to Appendix B to take the Self Check for this section.

Access the following online resources for more instruction and practice with solving simple equations.

. Solving One Step Equations By Addition and Subtraction (http:llopenstaxcollege.orglll245olveonestep)

. Solving One Step Equation by Adding/Subtracting Integers (Variable on Left)(http:llopenstaxcol lege,orglll24Solveoneleft)

. Solving One Step Equation by Adding/Subtracting lntegers (Variable on Right)(http:llopenstaxcoll ege.orgll/24solveonerig ht)

2.3 i solving equations using subtraction and addition properties … · 2.3.2 model the...

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