2.1 solutions of linear equations in one · pdf filesolutions of linear equations in one...

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Solutions of Linear Equations in One Variable2.1 OBJECTIVES

1. Identify a linear equation2. Combine like terms to solve an equation

We begin this chapter by considering one of the most important tools of mathematics—theequation. The ability to recognize and solve various types of equations and inequalities isprobably the most useful algebraic skill you will learn, and we will continue to build on themethods developed here throughout the remainder of the text. To start, let’s describe whatwe mean by an equation.

An equation is a mathematical statement in which two expressions represent the samequantity. An equation has three parts:

5x � 6 � 2x � 3

Left sidesignEquals

Right side

The equation simply says that the expression on the left and the expression on the rightrepresent the same quantity.

In this chapter, we will work with a particular kind of equation.

NOTE Linear equations arealso called first-degreeequations because the highestpower of the variable is thefirst power, or first degree.

NOTE We also say the solutionsatisfies the equation.

NOTE We use the questionmark over the symbol ofequality when we are checkingto see if the statement is true.

The solution of an equation in one variable is any number that will make the equation atrue statement. The solution set for such an equation is simply the set consisting of allsolutions.

� �

A linear equation in one variable is any equation that can be written in theform

ax � b � 0

in which a and b are any real numbers

a � 0

Definitions: Linear Equation in One Variable

Checking Solutions

Verify that �3 is a solution for the equation.

5x � 6 � 2x � 3

Replacing x with �3 gives

5(�3) � 6 � 2(�3) � 3

�15 � 6 � �6 � 3

�9 � �9 A true statement.

Example 1

2.1

SOLUTIONS TO LINEAR EQUATIONS IN ONE VARIABLE SECTION 2.1 51©

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Solving linear equations in one variable will require using equivalent equations.

C H E C K Y O U R S E L F 1

Verify that 7 is a solution for this equation.

5x � 15 � 2x � 6

NOTE You can easily verify thisby replacing x with �3 in eachequation.

NOTE Multiplying both sidesof an equation by the samenonzero quantity gives anequivalent equation.

NOTE Adding the samequantity to both sides of anequation gives an equivalentequation, which holds truewhether c is positive ornegative.

Two equations are equivalent if they have the same solution set.

Definitions: Equivalent Equations

For example, the three equations

5x � 5 � 2x � 4 3x � �9 x � �3

are all equivalent because they all have the same solution set, ��3�. Note that replacing xwith �3 will give a true statement in the third equation, but it is not as clear that �3 is a so-lution for the other two equations. This leads us to an equation-solving strategy of isolatingthe variable, as is the case in the equation x � �3.

To form equivalent equations that will lead to the solution of a linear equation, we needtwo properties of equations: addition and multiplication. The addition property is definedhere.

Recall that subtraction can always be defined in terms of addition, so

a � c � a � (�c)

The addition property also allows us to subtract the same quantity from both sides of anequation.

The multiplication property is defined here.

If a � b

then a � c � b � c

Rules and Properties: Addition Property of Equations

If a � b

then ac � bc when c � 0

Rules and Properties: Multiplication Property of Equations

It is also the case that division can be defined in terms of multiplication, so

c � 0

The multiplication property allows us to divide both sides of an equation by the same non-zero quantity.

a

c� a �

1

c

Applying the Properties of Equations

Solve for x.

3x � 5 � 4 (1)

We start by using the addition property to add 5 to both sides of the equation.

3x � 5 � 5 � 4 � 5

3x � 9 (2)

Now we want to get the x term alone on the left with a coefficient of 1 (we call thisisolating the x). To do this, we use the multiplication property and multiply both sides

by .

So, x � 3 (3)

In set notation, we write �3�, which represents the set of all solutions. No other valueof x makes the original equation true. We could also use set-builder notation. We write�x �x � 3�, which is read, “Every x such that x equals three.” We will use both notationsthroughout the text.

Because any application of the addition or multiplication properties leads to an equiva-lent equation, equations (1), (2), and (3) in Example 2 all have the same solution, 3.

To check this result, we can replace x with 3 in the original equation:

3(3) � 5 � 4

9 � 5 � 4

4 � 4 A true statement.

You may prefer a slightly different approach in the last step of the solution above. Fromequation (2),

3x � 9

The multiplication property can be used to divide both sides of the equation by 3. Then,

x � 3

Of course, the result is the same.

3x

3�

9

3

�1

3� 3�(x) � 3

1

3 (3x) �

1

3 (9)

1

3

52 CHAPTER 2 LINEAR EQUATIONS AND INEQUALITIES

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NOTE Why did we add 5? Weadded 5 because it is theopposite of �5, and theresulting equation will have thevariable term on the left andthe constant term on the right.

NOTE We choose because

is the reciprocal of 3 and

� 113

� 3

13

13

Example 2


Solve for x.

4x � 7 � 17

The steps involved in using the addition and multiplication properties to solve an equa-tion are the same if more terms are involved in an equation.


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Applying the Properties of Equations

Solve for x.

5x � 11 � 2x � 7

Our objective is to use the properties of equations to isolate x on one side of an equiva-lent equation. We begin by adding 11 to both sides.

5x � 11 � 11 � 2x � 7 � 11

5x � 2x � 4

We continue by adding �2x to (or subtracting 2x from) both sides. We can do this becauseof our addition property of equations.

5x � (�2x) � 2x � (�2x) � 4

We have now isolated the variable term on the left side of the equation.

3x � 4

In the final step, we multiply both sides by .

In set notation, we write . We leave it to you to check this result by substitution.�4

3x �

4

3

1

3 (3x) �

1

3 (4)

1

3

NOTE If you prefer, write

5x � 2x � 2x � 2x � 4

Again:

3x � 4

NOTE Notice the like terms onthe left and right sides of theequation.

NOTE This is the same asdividing both sides by 3. So

�

x �43

43

3x3

NOTE Adding 11 puts theconstant term on the right.

Example 3

Both sides of an equation should be simplified as much as possible before the additionand multiplication properties are applied. If like terms are involved on one side (or on bothsides) of an equation, they should be combined before an attempt is made to isolate thevariable. Example 4 illustrates this approach.


Solve for x.

7x � 12 � 2x � 9

Example 4

Applying the Properties of Equations with Like Terms

Solve for x.

8x � 2 � 3x � 8 � 3x � 2

Here we combine the like terms 8x and �3x on the left and the like terms 8 and 2 on theright as our first step. We then have

5x � 2 � 3x � 10

We can now solve as before.

5x � 2 � 2 � 3x � 10 � 2 Subtract 2 from both sides.

5x � 3x � 8

Then,

5x � 3x � 3x � 3x � 8 Subtract 3x from both sides.

2x � 8

� Divide both sides by 2.

x � 4 or �4�

The solution is 4, which can be checked by returning to the original equation.

8

2

2x

2


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NOTE Recall that to isolate thex, we must get x alone on theleft side with a coefficient of 1.


Solve for x.

7x � 3 � 5x � 10 � 4x � 3

If parentheses are involved on one or both sides of an equation, the parentheses shouldbe removed by applying the distributive property as the first step. Like terms should then becombined before an attempt is made to isolate the variable. Consider Example 5.

Applying the Properties of Equations with Parentheses

Solve for x.

x � 3(3x � 1) � 4(x � 2) � 4

First, apply the distributive property to remove the parentheses on the left and right sides.

x � 9x � 3 � 4x � 8 � 4

Combine like terms on each side of the equation.

10x � 3 � 4x � 12

Now, isolate variable x on the left side.

10x � 3 � 3 � 4x � 12 � 3 Add 3 to both sides.

10x � 4x � 15

10x � 4x � 4x � 4x � 15 Subtract 4x from both sides.

6x � 15

� Divide both sides by 6.

x � or

The solution is . Again, this can be checked by returning to the original equation.5

2

�5

25

2

15

6

6x

6

Example 5


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To solve an equation involving fractions, the first step is to multiply both sides of theequation by the least common multiple (LCM) of all denominators in the equation. Thiswill clear the equation of fractions, and we can proceed as before.


Solve for x.

x � 5(x � 2) � 3(3x � 2) � 18

NOTE The LCM of a set ofdenominators is also called thelowest common denominator(LCD).

NOTE The equation is nowcleared of fractions.

Applying the Properties of Equations with Fractions

Solve for x.

First, multiply each side by 6, the least common multiple of 2, 3, and 6.

Apply the distributive property.

Simplify.

Next, isolate the variable x on the left side.

3x � 4 � 5

3x � 9

x � 3 or �3�

The solution, 3, can be checked as before by returning to the original equation.

6�x

2� � 6�2

3� � 6�5

6�

6�x

2� � 6�2

3� � 6�5

6�

6�x

2�

2

3� � 6�5

6�

x

2�

2

3�

5

6

Example 6

3 2 1

1 1 1


Solve for x.

x

4�

4

5�

19

20

Be sure that the distributive property is applied properly so that every term of the equa-tion is multiplied by the LCM.

First, multiply each side by 10, the LCM of 5 and 2.

Apply the distributive property on the left. Reduce.

2(2x � 1) � 10 � 5x

4x � 2 � 10 � 5x

4x � 8 � 5x Next, isolate x. Here we isolate x on the right side.

8 � x or �8�

The solution for the original equation is 8.

10�2x � 1

5 � � 10(1) � 10�x

2�

10�2x � 1

5� 1� � 10�x

2�


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52


Solve for x.

3x � 1

4� 2 �

x � 1

3

Thus far, we have considered only equations of the form ax � b � 0, in which a � 0.If we allow the possibility that a � 0, two additional equation forms arise. The resultingequations can be classified into three types depending on the nature of their solutions.

1. An equation that is true for only particular values of the variable is called aconditional equation. Here the equation can be written in the form

ax � b � 0

in which a � 0. This case was illustrated in all our previous examples and exercises.

2. An equation that is true for all possible values of the variable is called an identity. Inthis case, both a and b are 0, so we get the equation 0 � 0. This will be the case ifboth sides of the equation reduce to the same expression (a true statement).

3. An equation that is never true, no matter what the value of the variable, is called acontradiction. For example, if a is 0 but b is nonzero, we end up with something like4 � 0. This will be the case if both sides of the equation reduce to a false statement.

Example 8 illustrates the second and third cases.

Example 7

Applying the Properties of Equations with Fractions

Solve for x.

2x � 1

5� 1 �

x

2


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Identities and Contradictions

(a) Solve for x.

2(x � 3) � 2x � �6

Apply the distributive property to remove the parentheses.

2x � 6 � 2x � �6

�6 � �6 A true statement.

Because the two sides reduce to the true statement �6 � �6, the original equation is anidentity, and the solution set is the set of all real numbers.

Example 8

NOTE See the definition of anidentity, above. By adding 6 toboth sides of this equation, wehave 0 � 0.

NOTE See the earlier definitionof a contradiction. Subtracting3 from both sides, we have0 � 1.

NOTE An algorithm is a step-by-step process for problemsolving.

An organized step-by-step procedure is the key to an effective equation-solving strategy.The following algorithm summarizes our work in this section and gives you guidance inapproaching the problems that follow.


Determine whether each of the following equations is a conditional equation, anidentity, or a contradiction.

(a) 2(x � 1) � 3 � x (b) 2(x � 1) � 3 � 2x � 1 (c) 2(x � 1) � 3 � 2x � 1

Step 1 Remove any grouping symbols by applying the distributive property.Step 2 Multiply both sides of the equation by the LCM of any denominators,

to clear the equation of fractions.Step 3 Combine any like terms that appear on either side of the equation.Step 4 Apply the addition property of equations to write an equivalent

equation with the variable term on one side of the equation and theconstant term on the other side.

Step 5 Apply the multiplication property of equations to write an equivalentequation with the variable isolated on one side of the equation.

Step 6 Check the solution in the original equation.

Step by Step: Solving Linear Equations in One Variable

(b) Solve for x.

3(x � 1) � 2x � x � 4

Again, apply the distributive property.

3x � 3 � 2x � x � 4

x � 3 � x � 4

3 � 4 A false statement.

Because the two sides reduce to the false statement 3 � 4, the original equation is a con-tradiction. There are no values of the variable that can satisfy the equation. The solution sethas nothing in it. We call this the empty set and write � � or �.

NOTE If the equation derivedin step 5 is always true, theoriginal equation was anidentity. If the equation isalways false, the originalequation was a contradiction.


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Evaluating Expressions Using a Calculator

Solve the following equation for x.

Following the steps of the algorithm, we get

Remove parentheses.

185x � 185 � 3.25 � 1650 � 159.44 � 500 Multiply by the LCM.

185x � 159.44 � 500 � 185 � 3.25 � 1650 Apply the addition property.

Isolate the variable.

Now, remembering to insert parentheses around the numerator, we use a calculator tosimplify the expression on the right.

x � 425.25 or �425.25�

x �159.44 � 500 � 185 � 3.25 � 1650

185

185x � 185 � 3.25 � 1650

500� 159.44

185(x � 3.25) � 1650

500� 159.44

Example 9

When you are solving an equation for which a calculator is recommended, it is ofteneasiest to do all calculations as the last step.


Solve the following equation for x.

2200(x � 17.5) � 1550

75� 2326

C H E C K Y O U R S E L F A N S W E R S

1. 5(7) � 15 � 2(7) � 6 2. �6� 3. 4. ��8� 5.35 � 15 � 14 � 6

20 � 20A true statement.

6. �7� 7. �5� 8. (a) Conditional; (b) contradiction; (c) identity 9. �62.5�

��2

3�3

5

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2.1

Name

Section Date

ANSWERS

1. 2.

3. 4.

5. 6.

7. 8.

9. 10.

11. 12.

13.

14.

15.

16.

17.

18.

19.

20.

21.

22.

23.

24.

25.

26.

27.

28.

59

Exercises

In exercises 1 to 14, solve each equation, and check your results. Express each answer inset notation.

1. 5x � 8 � 17 2. 4x � 9 � �11

3. 8 � 7x � �41 4. �7 � 4x � 21

5. 7x � 5 � 6x � 6 6. 9x � 4 � 8x � 3

7. 8x � 4 � 3x � 24 8. 5x � 2 � 2x � 5

9. 7x � 4 � 2x � 26 10. 11x � 3 � 4x � 31

11. 4x � 3 � 1 � 2x 12. 8x � 5 � �19 � 4x

13. 2x � 8 � 7x � 37 14. 3x � 5 � 9x � 22

In exercises 15 to 32, simplify and then solve each equation. Express your answer in setnotation.

15. 5x � 2 � x � 9 � 3x � 10 16. 5x � 5 � x � �7 � x � 2

17. 7x � 3 � 4x � 5 � 5x � 13 18. 8x � 3 � 6x � 7 � 5x � 17

19. 5x � 3(x � 6) 20. 2(x � 15) � 7x

21. 5(8 � x) � 3x 22. 7x � 7(6 � x)

23. 2(2x � 1) � 3(x � 1) 24. 3(3x � 1) � 4(3x � 1)

25. 8x � 3(2x � 4) � 17 26. 7x � 4(3x � 4) � 9

27. 7(3x � 4) � 8(2x � 5) � 13 28. �4(2x � 1) � 3(3x � 1) � 9

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29. 9 � 4(3x � 1) � 3(6 � 3x) � 9

30. 13 � 4(5x � 1) � 3(7 � 5x) � 15

31. 5 � 2[x � 2(x � 1)] � 55 � 4[x � 3(x � 2)]

32. 7 � 5[x � 3(x � 2)] � 25 � 2[x � 2(x � 3)]

In exercises 33 to 46, clear fractions and then solve each equation. Express your answer inset notation.

33. 34.

35. 36.

37. 38.

39. 40.

41. 42.

43. 44.

45. 0.5x � 6 � 0.2x 46. 0.7x � 7 � 0.3x � 5

In exercises 47 to 56, classify each equation as a conditional equation, an identity, or acontradiction.

47. 3(x � 1) � 2x � 3 48. 2(x � 3) � 2x � 6

49. 3(x � 1) � 3x � 3 50. 2(x � 3) � x � 5

3x

5�

3x � 1

2�

11

10

2x � 3

5�

2x � 1

3�

8

15

6x � 1

5�

2x

3� 3

5x � 3

4� 2 �

x

3

x

6�

3

4�

x � 1

4

x

5�

x � 7

3�

1

3

5x

6�

2x

3�

5

6

2x

3�

x

4�

5

2

x

6�

x

8� 1

x

6�

x

5� 11

3x

4�

1

4� 4

2x

3�

5

3� 3

ANSWERS

29.

30.

31.

32.

33.

34.

35.

36.

37.

38.

39.

40.

41.

42.

43.

44.

45.

46.

47.

48.

49.

50.

60

51. 3(x � 1) � 3x � 3 52. 2(x � 3) � 3x � 5

53. 3x � (x � 3) � 2(x � 1) � 2 54. 5x � (x � 4) � 4(x � 2) � 4

55. 56.

In exercises 57 to 60, use a calculator to solve the given equations for x. Round youranswer to two decimal places and use set notation.

57. 58.

59. 60.

61. What is the common characteristic of equivalent equations?

62. What is meant by a solution to a linear equation?

63. Define (a) identity and (b) contradiction.

64. Why does the multiplication property of equation not include multiplying both sidesof the equation by 0?

Label exercises 65 to 70 true or false.

65. Adding the same value to both sides of an equation creates an equivalent equation.

66. Multiplying both sides of an equation by 0 creates an equivalent equation.

67. To clear an equation of fractions, we multiply both sides by the GCF of thedenominator.

68. The multiplication property of equations allows us to divide both sides by the samenonzero quantity.

�15.25x � 12(2x � 11.23)

�15.6� 8.4

�23x � 14(x � 9.75)

23.46� 15.75

47(x � 3.15) � 263

315� 11

63(x � 2.45) � 325

200� 3

3x

4�

2x

3�

x

6

x

2�

x

3�

x

6

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sANSWERS

51.

52.

53.

54.

55.

56.

57.

58.

59.

60.

61.

62.

63.

64.

65.

66.

67.

68.

61

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69. Some equations have more than one solution.

70. No matter what value is substituted for x, the expressions on either side of the equalssign have the same value.

Answers

1. �5� 3. �7� 5. �11� 7. ��4� 9. �6� 11. 13. �9�

15. �7� 17. 19. ��9� 21. �5� 23. �5� 25.

27. �5� 29. 31. ��13� 33. �7� 35. �30� 37. �6�

39. �15� 41. �3� 43. 45. �20� 47. Conditional

49. Contradiction 51. Identity 53. Contradiction 55. Identity57. �6.82� 59. ��6.30� 61. 63. 65. True67. False 69. True

�3

2��

4

3�5

2�5

2�2

3

ANSWERS

69.

70.

62

2.1 solutions of linear equations in one · pdf filesolutions of linear equations in one...

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