Lesson 4: Solving a Linear Equation

8•4 Lesson 4

Lesson 4: Solving a Linear Equation


Student Outcomes

Students extend the use of the properties of equality to solve linear equations having rational coefficients.

Classwork Concept Development (13 minutes)

To solve an equation means to find all of the numbers 𝑥𝑥, if they exist, so that the given equation is true.

In some cases, some simple guesswork can lead us to a solution. For example, consider the following equation:

4𝑥𝑥 + 1 = 13.

What number 𝑥𝑥 would make this equation true? That is, what value of 𝑥𝑥 would make the left side equal to the right side? (Give students a moment to guess a solution.) When 𝑥𝑥 = 3, we get a true statement. The left side of the equal sign is equal to 13, and so is the right

side of the equal sign.

In other cases, guessing the correct answer is not so easy. Consider the following equation:

3(4𝑥𝑥 − 9) + 10 = 15𝑥𝑥 + 2 + 7𝑥𝑥.

Can you guess a number for 𝑥𝑥 that would make this equation true? (Give students a minute to guess.)

Guessing is not always an efficient strategy for solving equations. In the last example, there are several terms in each of the linear expressions comprising the equation. This makes it more difficult to easily guess a solution. For this reason, we want to use what we know about the properties of equality to transform equations into equations with fewer terms.

The ultimate goal of solving any equation is to get it into the form of 𝑥𝑥 (or whatever symbol is being used in the equation) equal to a constant.

Complete the activity described below to remind students of the properties of equality, and then proceed with the discussion that follows.

Give students the equation 4 + 1 = 7 − 2, and ask them the following questions:

1. Is this equation true?

2. Perform each of the following operations, and state whether or not the equation is still true:

a. Add three to both sides of the equal sign.

b. Add three to the left side of the equal sign, and add two to the right side of the equal sign.

c. Subtract six from both sides of the equal sign.

d. Subtract three from one side of the equal sign, and subtract three from the other side of the equal sign.

e. Multiply both sides of the equal sign by ten.

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8•4 Lesson 4


f. Multiply the left side of the equation by ten and the right side of the equation by four.

g. Divide both sides of the equation by two.

h. Divide the left side of the equation by two and the right side of the equation by five.

3. What do you notice? Describe any patterns you see.

There are four properties of equality that allow us to transform an equation into the form we want. If 𝐴𝐴, 𝐵𝐵,

and 𝐶𝐶 are any rational numbers, then:

- If 𝐴𝐴 = 𝐵𝐵, then 𝐴𝐴 + 𝐶𝐶 = 𝐵𝐵 + 𝐶𝐶. - If 𝐴𝐴 = 𝐵𝐵, then 𝐴𝐴 − 𝐶𝐶 = 𝐵𝐵 − 𝐶𝐶. - If 𝐴𝐴 = 𝐵𝐵, then 𝐴𝐴 ⋅ 𝐶𝐶 = 𝐵𝐵 ⋅ 𝐶𝐶.

- If 𝐴𝐴 = 𝐵𝐵, then 𝐴𝐴𝐶𝐶

=𝐵𝐵𝐶𝐶

, where 𝐶𝐶 is not equal to zero.

All four of the properties require us to start off with 𝐴𝐴 = 𝐵𝐵. That is, we have to assume that a given equation has an expression on the left side that is equal to the expression on the right side. Working under that assumption, each time we use one of the properties of equality, we are transforming the equation into another equation that is also true; that is, the left side equals the right side.

Example 1 (3 minutes)

Solve the linear equation 2𝑥𝑥 − 3 = 4𝑥𝑥 for the number 𝑥𝑥. Examine the properties of equality. Choose “something” to add, subtract, multiply, or divide on both sides of

the equation.

Validate the use of the properties of equality by having students share their thoughts. Then, discuss the “best” choice for the first step in solving the equation with the points below. Be sure to remind students throughout this and the other examples that the goal is to get 𝑥𝑥 equal to a constant; therefore, the “best” choice is one that gets them to that goal most efficiently.

First, we must assume that there is a number 𝑥𝑥 that makes the equation true. Working under that assumption, when we use the property, if 𝐴𝐴 = 𝐵𝐵, then 𝐴𝐴 − 𝐶𝐶 = 𝐵𝐵 − 𝐶𝐶, we get an equation that is also true.

2𝑥𝑥 − 3 = 4𝑥𝑥 2𝑥𝑥 − 2𝑥𝑥 − 3 = 4𝑥𝑥 − 2𝑥𝑥

Now, using the distributive property, we get another set of equations that is also true.

(2 − 2)𝑥𝑥 − 3 = (4 − 2)𝑥𝑥 0𝑥𝑥 − 3 = 2𝑥𝑥

−3 = 2𝑥𝑥

Using another property, if 𝐴𝐴 = 𝐵𝐵, then 𝐴𝐴𝐶𝐶

=𝐵𝐵𝐶𝐶

, we get another equation that is also true.

−32

=2𝑥𝑥2

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8•4 Lesson 4


After simplifying the fraction 22

, we have

−32

= 𝑥𝑥,

which is also true.

The last step is to check to see if 𝑥𝑥 = − 32 satisfies the equation 2𝑥𝑥 − 3 = 4𝑥𝑥.

The left side of the equation is equal to 2 ⋅ �− 32� − 3 = −3 − 3 = −6.

The right side of the equation is equal to 4 ⋅ �− 32� = 2 ⋅ (−3) = −6.

Since the left side equals the right side, we know we have found the number 𝑥𝑥 that solves the equation 2𝑥𝑥 − 3 = 4𝑥𝑥.


Solve the linear equation 35𝑥𝑥 − 21 = 15. Keep in mind that our goal is to transform the equation so that it is

in the form of 𝑥𝑥 equal to a constant. If we assume that the equation is true for some number 𝑥𝑥, which property should we use to help us reach our goal, and how should we use it?

Again, provide students time to decide which property is “best” to use first.

We should use the property if 𝐴𝐴 = 𝐵𝐵, then 𝐴𝐴 + 𝐶𝐶 = 𝐵𝐵 + 𝐶𝐶, where the number 𝐶𝐶 is 21.

Note that if students suggest that they subtract 15 from both sides (i.e., where 𝐶𝐶 is −15), remind them that they want the form of 𝑥𝑥 equal to a constant. Subtracting 15 from both sides of the equal sign puts the 𝑥𝑥 and all of the constants on the same side of the equal sign. There is nothing mathematically incorrect about subtracting 15, but it does not get them any closer to reaching the goal.

If we use 𝐴𝐴 + 𝐶𝐶 = 𝐵𝐵 + 𝐶𝐶, then we have the true statement:

35𝑥𝑥 − 21 + 21 = 15 + 21

and

35𝑥𝑥 = 36.

Which property should we use to reach our goal, and how should we use it?

We should use the property if 𝐴𝐴 = 𝐵𝐵, then 𝐴𝐴 ⋅ 𝐶𝐶 = 𝐵𝐵 ⋅ 𝐶𝐶, where 𝐶𝐶 is 53

.

If we use 𝐴𝐴 ⋅ 𝐶𝐶 = 𝐵𝐵 ⋅ 𝐶𝐶, then we have the true statement:

35𝑥𝑥 ⋅

53

= 36 ⋅53

,

and by the commutative property and the cancellation law, we have

𝑥𝑥 = 12 ⋅ 5 = 60.

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8•4 Lesson 4


Note to Teacher: There are many ways to solve this equation. Any of the actions listed below are acceptable. In fact, a student could say, “Add 100 to both sides of the equal sign,” and that, too, would be an acceptable action. It may not lead directly to the answer, but it is still an action that would make a mathematically correct statement. Make it clear that it does not matter which option students choose or in which order; what matters is that they use the properties of equality to make true statements that lead to a solution in the form of 𝑥𝑥 equal to a constant.

Does 𝑥𝑥 = 60 satisfy the equation 35𝑥𝑥 − 21 = 15?

Yes, because the left side of the equation is equal to 1805 − 21 = 36 − 21 = 15. Since the right side is

also 15, then we know that 60 is a solution to 35 𝑥𝑥 − 21 = 15.


The properties of equality are not the only properties we can use with equations. What other properties do we know that could make solving an equation more efficient? We know the distributive property, which allows us to expand and simplify expressions.

We know the commutative and associative properties, which allow us to rearrange and group terms within expressions.

Now we will solve the linear equation 15𝑥𝑥 + 13 + 𝑥𝑥 = 1 − 9𝑥𝑥 + 22. Is there anything we can do to the linear

expression on the left side to transform it into an expression with fewer terms?

Yes. We can use the commutative and distributive properties:

15𝑥𝑥 + 13 + 𝑥𝑥 =

15𝑥𝑥 + 𝑥𝑥 + 13

= �15

+ 1� 𝑥𝑥 + 13

=65𝑥𝑥 + 13.

Is there anything we can do to the linear expression on the right side to transform it into an expression with fewer terms? Yes. We can use the commutative property:

1 − 9𝑥𝑥 + 22 = 1 + 22 − 9𝑥𝑥 = 23 − 9𝑥𝑥.

Now we have the equation 65 𝑥𝑥 + 13 = 23 − 9𝑥𝑥. What should we do now to solve the equation?

Students should come up with the following four responses as to what should be done next. A case can be made for each of them being the “best” move. In this case, each move gets them one step closer to the goal of having the solution in the form of 𝑥𝑥 equal to a constant. Select one option, and move forward with solving the equation (the notes that follow align to the first choice, subtracting 13 from both sides of the equal sign).

We should subtract 13 from both sides of the equal sign. We should subtract 23 from both sides of the equal sign. We should add 9𝑥𝑥 to both sides of the equal sign.

We should subtract 65𝑥𝑥 from both sides of the equal sign.

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8•4 Lesson 4


Let’s choose to subtract 13 from both sides of the equal sign. Though all options were generally equal with respect to being the “best” first step, I chose this one because when I subtract 13 on both sides, the value of the constant on the left side is positive. I prefer to work with positive numbers. Then we have

65𝑥𝑥 + 13 − 13 = 23 − 13 − 9𝑥𝑥

65𝑥𝑥 = 10 − 9𝑥𝑥

What should we do next? Why?

We should add 9𝑥𝑥 to both sides of the equal sign. We want our solution in the form of 𝑥𝑥 equal to a constant, and this move puts all terms with an 𝑥𝑥 on the same side of the equal sign.

Adding 9𝑥𝑥 to both sides of the equal sign, we have

65𝑥𝑥 + 9𝑥𝑥 = 10 − 9𝑥𝑥 + 9𝑥𝑥

515𝑥𝑥 = 10.

What do we need to do now?

We should multiply 551

on both sides of the equal sign.

Then we have

515𝑥𝑥 ⋅

551

= 10 ⋅5

51.

By the commutative property and the fact that 551

⋅515

= 1, we have

𝑥𝑥 =5051

.

All of the work we did is only valid if our assumption that 15𝑥𝑥 + 13 + 𝑥𝑥 = 1 − 9𝑥𝑥 + 22 is a true statement.

Therefore, check to see if 5051

makes the original equation a true statement.

15𝑥𝑥 + 13 + 𝑥𝑥 = 1 − 9𝑥𝑥 + 22

15�

5051� + 13 +

5051

= 1 − 9 �5051� + 22

65�

5051� + 13 = 23 − 9 �

5051�

300255

+ 13 = 23 −45051

3615255

=72351

72351

=72351

Since both sides of our equation equal 72351

, then we know that 5051

is a solution of the equation.

Note to Teacher: There are still options. If students say they should

subtract 65𝑥𝑥 from both sides of

the equal sign, remind them of the goal of obtaining the 𝑥𝑥 equal to a constant.

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8•4 Lesson 4


Problem Set For each problem, show your work, and check that your solution is correct.

1. Solve the linear equation 𝑥𝑥 + 4 + 3𝑥𝑥 = 72. State the property that justifies your first step and why you chose it.

2. Solve the linear equation 𝑥𝑥 + 3 + 𝑥𝑥 − 8 + 𝑥𝑥 = 55. State the property that justifies your first step and why you chose it.

3. Solve the linear equation 12𝑥𝑥 + 10 =

14𝑥𝑥 + 54. State the property that justifies your first step and why you chose

it.

4. Solve the linear equation 14𝑥𝑥 + 18 = 𝑥𝑥. State the property that justifies your first step and why you chose it.

5. Solve the linear equation 17 − 𝑥𝑥 = 13 ⋅ 15 + 6. State the property that justifies your first step and why you chose it.

6. Solve the linear equation 𝑥𝑥+𝑥𝑥+2

4= 189.5. State the property that justifies your first step and why you chose it.

Lesson Summary

The properties of equality, shown below, are used to transform equations into simpler forms. If 𝐴𝐴,𝐵𝐵,𝐶𝐶 are rational numbers, then:

If 𝐴𝐴 = 𝐵𝐵, then 𝐴𝐴 + 𝐶𝐶 = 𝐵𝐵 + 𝐶𝐶. Addition property of equality

If 𝐴𝐴 = 𝐵𝐵, then 𝐴𝐴 − 𝐶𝐶 = 𝐵𝐵 − 𝐶𝐶. Subtraction property of equality

If 𝐴𝐴 = 𝐵𝐵, then 𝐴𝐴 ⋅ 𝐶𝐶 = 𝐵𝐵 ⋅ 𝐶𝐶. Multiplication property of equality

If 𝐴𝐴 = 𝐵𝐵, then 𝐴𝐴𝐶𝐶

=𝐵𝐵𝐶𝐶

, where 𝐶𝐶 is not equal to zero. Division property of equality

To solve an equation, transform the equation until you get to the form of 𝑥𝑥 equal to a constant (𝑥𝑥 = 5, for example).

A STORY OF RATIOS

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S.12




Copyright © by Holt, Rinehart and Winston. 54 Holt MathematicsAll rights reserved.

Name Date Class

ReteachSolving Equations by Adding or Subtracting 1-7

LESSON

To solve an addition equation, use subtraction.

Solve x � 5 � 12.

x � 7

Tell what you would add or subtract to solve the equation.

1. a � 2 � 7 2. x � 2 � 4 3. 6 � y � 9 4. 34 � b � 13

5. 21 � z � 9 6. 14 � r � 20 7. 18 � d � 11 8. 6 � 5 � p

Complete to solve the equation. In Exercises 9–11 check the solution.

9. 10. 11.

x � � z � � � � n

x � z � � n

Check: x � 2 � 15 Check: z � 6 � 14 Check: 7 � 4 � n

� 2 � 15 � 6 � 14 7 � 4 �

� 15 � � 14 � 7 � �

12. 13. 14.

t � � a � � � � b

t � a � � b192011

019200110

28 � 9 � ba � 7 � 13t � 5 � 16

71415

3817

3817

0380170

7 � 4 � nz � 6 � 14x � 2 � 15� 2 � 2

subtract 5add 11subtract 14add 9

subtract 13subtract 6add 2subtract 2

x � 5 � 12� 5 � 5

To solve a subtraction equation, use addition.

Solve 9 � w � 3.

12 � w

9 � w � 3� 3 � 3

�6 �6

�5 �5 �7 �7

�4 �4

�9 �9

Subtract thenumber added to the variable.

Add the numbersubtracted from the variable.

MSM07G8_RESBK_Ch01_051_058.pe 1/2/06 12:05 PM Page 54

Copyright © by Holt, Rinehart and Winston. 62 Holt MathematicsAll rights reserved.

Name Date Class

ReteachSolving Equations by Multiplying or Dividing 1-8

LESSON

When an equation has two operations, undo addition or subtractionfirst. Then undo multiplication or division.Solve 2x � 11 � 35. 2x � 11 � 35

� 11 �11 Undo the addition.2x � 24

�22x� � �

224� Undo the multiplication.

x � 12

Tell what number you would multiply or divide by to solve theequation.

1. 5a � 60 2. �6x

� � 12 3. 144 � 12f

Solve.

4. 6x � 42 5. �a3

� � 9 6. 25 � �5k

�

�66x� � �4

62� • �

a3

� � 9 • • 25 � �5k

� •

x � a � � k

7. 2x � 3 � 11 8. �b4

� � 5 � 6 9. 5t � 9 � 36

�22x� � �

82

� • �b4

� � • �55t� � �

455�

x � b � t � 944

414

�9�9�5�5�3�3

125277

5577

Divid12.Mult6.Divide .

To solve a multiplication equation, usedivision.

Solve 3x � 24.

3x � 24

�33x� � �

234�

x � 8

To solve a division equation, usemultiplication.

Solve �4x

� � 20.

�4x

� � 20

4 • �4x

� � 20 • 4

x � 80

MSM07G8_RESBK_Ch01_059_066.pe 1/2/06 12:04 PM Page 62

Solving One-Step Equations

Directions: Solve each one-step equation. Use your answer to navigate through the maze.Show your work.

𝒙 + 𝟑 = 𝟏𝟐

𝒙 = 𝟗

𝟏𝟓 + 𝒙 = 𝟏𝟕

𝒙 = 𝟏𝟓

𝒙 − 𝟑𝟐 = 𝟏𝟐

𝒙 = 𝟒𝟖

𝟑𝟔 − 𝒙 = 𝟐𝟔

𝒙 = 𝟐

𝒙 = −𝟖

𝟒𝒙 = 𝟏𝟐 𝟕𝒙 = 𝟒𝟐

𝟐𝟓

𝒙= −𝟏

𝒙 = 𝟗

𝒙 = 𝟏𝟖𝟎

−𝟕𝟐

𝒙= 𝟗

𝒙

𝟒= 𝟔

𝒙 = 𝟐𝟒

−𝟑𝟎

𝒙= −𝟔

𝟏𝟒 − 𝒙 = 𝟔

𝒙 =𝟔

𝟒

𝒙 = 𝟐𝟎

𝒙 = 𝟐𝟏

−𝟐𝒙 = −𝟒𝟐

−𝟐 + 𝒙 = 𝟏𝟓 𝒙

−𝟒= −𝟕

𝒙 = −𝟕

𝒙 − 𝟏𝟑 = 𝟐𝟑

𝒙 = 𝟐𝟗𝟒

𝒙 = 𝟔

𝒙 = 𝟓

𝒙 = 𝟐𝟖 𝒙 = 𝟏𝟑

𝒙 = 𝟏𝟕

𝒙 = 𝟓𝟎

𝒙 = 𝟐𝟓 𝒙 = −𝟑

𝒙 = −𝟓

𝒙 = 𝟏𝟎

𝒙 = 𝟖

Lesson 4: Solving a Linear Equation

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