# 4.6.1: Solving Equations Involving Integers

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We return to solving equations involving integers, only this time the equations will be a bit more advanced, requiring the use of the distributive property and skill at combining like terms. Let’s begin.

Example 1

Solve for x: 7x − 11x = 12.

Solution

Combine like terms.

\begin{aligned} 7x-11x=12 ~ & \textcolor{red}{ \text{ Original equation.}} \\ =4x=12 ~ & \textcolor{red}{ \text{ Combine like terms: } 7x-11x=-4x.} \end{aligned}\nonumber

To undo the effect of multiplying by −4, divide both sides of the last equation by −4.

\begin{aligned} \frac{-4x}{-4} = \frac{12}{-4} ~ & \textcolor{red}{ \text{ Divide both sides by } -4.} \\ x = -3 ~ & \textcolor{red}{ \text{ Simplify: } 12/(-4) = -3.} \end{aligned}\nonumber

Check

Substitute −3 for x in the original equation.

\begin{aligned} 7x-11x = 12 ~ & \textcolor{red}{ \text{ Original equation.}} \\ 7(-3)-11(-3)=12 ~ & \textcolor{red}{ \text{ Substitute } -3 \text{ for } x.} \\ -21 + 33 = 12 ~ & \textcolor{red}{ \text{ On the left, multiply first.}} \\ 12 = 12 ~ & \textcolor{red}{ \text{ On the left, add.}} \end{aligned}\nonumber

Because the last line of the check is a true statement, −3 is a solution of the original equation.

Exercise

Solve for x: −6x − 5x = 22.

x = −2

Example 2

Solve for x: 12 = 5x − (4 + x).

Solution

To take the negative of a sum, negate each term in the sum (change each term to its opposite). Thus, −(4 + x) = −4 − x.

\begin{aligned} 12 = 5x - (4+x) ~ & \textcolor{red}{ \text{ Original equation.}} \\ 12 = 5x - 4 - x ~ & \textcolor{red}{-(4+x)=-4-x.} \\ 12 = 4x-4 ~ & \textcolor{red}{ \text{ Combine like terms: } 5x - x = 4x.} \end{aligned}\nonumber

To undo the effect of subtracting 4, add 4 to both sides of the last equation.

\begin{aligned} 12 + 4 = 4x-4+4 ~ & \textcolor{red}{ \text{ Add 4 to both sides.}} \\ 16 = 4x ~ & \textcolor{red}{ \text{ Simplify both sides.}} \end{aligned}\nonumber

To undo the effect of multiplying by 4, divide both sides of the last equation by 4.

\begin{aligned} \frac{16}{4} = \frac{4x}{4} ~ & \textcolor{red}{ \text{ Divide both sides by 4.}} \\ 4 = x ~ & \textcolor{red}{ \text{ Simplify: 16/4=4.} \end{aligned}] Check Substitute 4 for x in the original equation. \[ \begin{aligned} 12 = 5x - (4+x) ~ & \textcolor{red}{ \text{ Original equation.}} \\ 12 = 5(4) - (4+4) ~ & \textcolor{red}{ \text{ Substitute 4 for } x.} \\ 12 = 20 - 8 ~ & \textcolor{red}{ \text{ On the right, 5(4) = 20 and evaluate}} \\ ~ & \textcolor{red}{ \text{ parentheses: } 4+4=8.} \\ 12 = 12 ~ & \textcolor{red}{ \text{ Simplify.}} \end{aligned}\nonumber

Because the last line of the check is a true statement, 4 is a solution of the original equation.

Exercise

Solve for x: 11 = 3x − (1 − x)

x = 3

## Variables on Both Sides

Variables can occur on both sides of the equation.

Goal

Isolate the terms containing the variable you are solving for on one side of the equation.

Example 3

Solve for x: 5x = 3x − 18.

Solution

To isolate the variables on one side of the equation, subtract 3x from both sides of the equation and simplify.

\begin{aligned} 5x = 3x-18 ~ & \textcolor{red}{ \text{ Original equation.}} \\ 5x-3x=3x-18-3x ~ & \textcolor{red}{ \text{ Subtract } 3x \text{ from both sides.}} \\ 2x = -18 ~ & \textcolor{red}{ \text{ Combine like terms: } 5x - 3x = 2x } \\ ~ & \textcolor{red}{ \text{ and } 3x - 3x = 0.} \end{aligned}\nonumber

Note that the variable is now isolated on the left-hand side of the equation. To undo the effect of multiplying by 2, divide both sides of the last equation by 2.

\begin{aligned} \frac{2x}{2} = \frac{-18}{2} ~ & \textcolor{red}{ \text{ Divide both sides by 2.}} \\ x = -9 ~ & \textcolor{red}{ \text{ Simplify: } -18/2 = =-9.} \end{aligned}\nonumber

Check

Substitute −9 for x in the original equation.

\begin{aligned} 5x = 3x - 18 ~ & \textcolor{red}{ \text{ Original equation.}} \\ 5(-9) = 3(-9)-18 ~ & \textcolor{red}{ \text{ Substitute } -9 \text{ for } x.} \\ -45 = -27-18 ~ & \textcolor{red}{ \text{ Multiply first on both sides.}} \\ -45 = -45 ~ & \textcolor{red}{ \text{ Subtract on the right: } -27 - 18 = -45.} \end{aligned}\nonumber

Because the last line of the check is a true statement, −9 is a solution of the original equation.

Exercise

Solve for x: 4x − 3 = x

x = 1

Example 4

Solve for x: 5x =3+6x.

Solution

To isolate the variables on one side of the equation, subtract 6x from both sides of the equation and simplify.

\begin{aligned} 5x = 3+6x ~ & \textcolor{red}{ \text{ Original equation.}} \\ 5x - 6x = 3+6x - 6x ~ & \textcolor{red}{ \text{ Subtract } 6x \text{ from both sides.}} \\ -x=3 ~ & \textcolor{red}{ \text{ Combine like terms: } 5x - 6x = -x} \\ ~ & \textcolor{red}{ \text{ and } 6x - 6x = 0.} \end{aligned}\nonumber

Note that the variable is now isolated on the left-hand side of the equation.

There are a couple of ways we can finish this solution. Remember, −x is the same as (−1)x, so we could undo the effects of multiplying by −1 by dividing both sides of the equation by −1. Multiplying both sides of the equation by −1 will work equally well. But perhaps the easiest way to proceed is to simply negate both sides of the equation.

\begin{aligned} -(-x) = -3 ~ & \textcolor{red}{ \text{ Negate both sides.}} \\ x = -3 ~ & \textcolor{red}{ \text{ Simplify: } -(-x) = x.} \end{aligned}\nonumber

Check

Substitute −3 for x in the original equation.

\begin{aligned} 5x = 3+6x ~ & \textcolor{red}{ \text{ Original equation.}} \\ 5(-3) = 3+6(-3) ~ & \textcolor{red}{ \text{ Substitute } -3 \text{ for } x.} \\ -15 = 3-18 ~ & \textcolor{red}{ \text{ Multiply first on both sides.}} \\ -15 = -15 ~ & \textcolor{red}{ \text{ Subtract on the right: } 3 -8 = -15.} \end{aligned}\nonumber

Because the last line of the check is a true statement, −3 is a solution of the original equation.

Exercise

Solve for x: 7x = 18 + 9x

x = −9

Dealing with −x.

If your equation has the form

x = c,

where c is some integer, note that this is equivalent to the equation (−1)x = c. Therefore, dividing both sides by −1 will produce a solution for x. Multiplying both sides by −1 works equally well. However, perhaps the easiest thing to do is negate each side, producing

−(−x) = −c, which is equivalent to x = −c.

Example 5

Solve for x: 6x − 5 = 12x + 19.

Solution

To isolate the variables on one side of the equation, subtract 12x from both sides of the equation and simplify.

\begin{aligned} 6x-5 = 12x+19 ~ & \textcolor{red}{ \text{ Original equation.}} \\ 6x-5-12x = 12x + 19 - 12x ~ & \textcolor{red}{ \text{ Subtract } 12x \text{ from both sides.}} \\ -6x - 5 = 19 ~ & \textcolor{red}{ \text{ Combine like terms: } 6x-12x=-6x} \\ ~ & \textcolor{red}{ \text{ and } 12x-12x = 0.} \end{aligned}\nonumber

Note that the variable is now isolated on the left-hand side of the equation. Next, to “undo” subtracting 5, add 5 to both sides of the equation.

\begin{aligned} -6x -5+5 = 19+5 ~ & \textcolor{red}{ \text{ Add 5 to both sides.}} \\ -6x = 24 ~ & \textcolor{red}{ \text{ Simplify: } -5+5=0 \text{ and } 19+5 = 24.} \end{aligned}\nonumber

Finally, to “undo” multiplying by −6, divide both sides of the equation by −6.

\begin{aligned} \frac{-6x}{-6} = \frac{24}{-6} ~ & \textcolor{red}{ \text{ Divide both sides by } -6.} \\ x = -4 ~ & \textcolor{red}{ \text{ Simplify: } 24/(-6)=-4.} \end{aligned}\nonumber

Check. Substitute −4 for x in the original equation

\begin{aligned} 6x-5=12x+19 ~ & \textcolor{red}{ Original equation.}} \\ 6(-4)-5 = 12(-4)+19 ~ & \textcolor{red}{ \text{ Substitute } -4 \text{ for } x.} \\ -24-5 = -48 + 19 ~ & \textcolor{red}{ \text{ Multiply first on both sides.}} \\ -29 = -29 ~ & \textcolor{red}{ Add: } -24 - 5 = -29 \text{ and } -48 + 19 = -29.} \end{aligned}\nonumber

Because the last line of the check is a true statement, −4 is a solution of the original equation.

Exercise

Solve for x: 2x + 3 = 18 − 3x

x = 3

Example 6

Solve for x: 2(3x + 2) − 3(4 − x) = x + 8.

Solution

Use the distributive property to remove parentheses on the left-hand side of the equation.

\begin{aligned} 2(3x+2)-3(4-x)=x+8 ~ & \textcolor{red}{ \text{ Original equation.}} \\ 6x+4 - 12 + 3x = x + 8 ~ & \textcolor{red}{ \text{ Use the distributive property.}} \\ 9x - 8 = x + 8 ~ & \textcolor{red}{ \text{ Combine like terms: } 6x + 3x = 9x} \\ ~ & \textcolor{red}{ \text{ and } 4-12 = -8.} \end{aligned}\nonumber

Isolate the variables on the left by subtracting x from both sides of the equation

\begin{aligned} 9x-8 -x = x + 8 - x ~ & \textcolor{red}{ \text{ Subtract } x \text{ from both sides.}} \\ 8x - 8 = 8 ~ & \textcolor{red}{ \text{ Combine like terms: } 9x-x = 8x} \\ ~ & \textcolor{red}{ \text{ and } x-x=0.} \end{aligned}\nonumber

Note that the variable is now isolated on the left-hand side of the equation. Next, to “undo” subtracting 8, add 8 to both sides of the equation.

\begin{aligned} 8x-8+8=8+8 ~ & \textcolor{red}{ \text{ Add 8 to both sides.}} \\ 8x = 16 ~ & \textcolor{red}{ \text{ Simplify: } -8+8=0 \text{ and } 8+8 = 16.} \end{aligned}\nonumber

Finally, to “undo” multiplying by 8, divide both sides of the equation by 8.

\begin{aligned} \frac{8x}{8} = \frac{16}{8} ~ & \textcolor{red}{ \text{ Divide both sides by 8.}} \\ x = 2 ~ & \textcolor{red}{ \text{ Simplify: } 16/8=2.} \end{aligned}\nonumber

Check. Substitute 2 for x in the original equation.

\begin{aligned} \end{aligned}\nonumber

Because the last line of the check is a true statement, 2 is a solution of the original equation.

Exercise

Solve for x: 3(2x − 4) − 2(5 − x) = 18

x = 5

## Exercises

In Exercises 1-16, solve the equation.

1. −9x + x = −8

2. 4x − 5x = −3

3. −4=3x − 4x

4. −6 = −5x + 7x

5. 27x + 51 = −84

6. −20x + 46 = 26

7. 9=5x + 9 − 6x

8. −6 = x + 3 − 4x

9. 0= −18x + 18

10. 0= −x + 71

11. 41 = 28x + 97

12. −65 = −x − 35

13. 8x − 8 − 9x = −3

14. 6x + 7 − 9x = 4

15. −85x + 85 = 0

16. 17x − 17 = 0

In Exercises 17-34, solve the equation.

17. −6x = −5x − 9

18. −5x = −3x − 2

19. 6x − 7=5x

20. 3x +8= −5x

21. 4x − 3=5x − 1

22. x − 2=9x − 2

23. −3x +5=3x − 1

24. −5x +9= −4x − 3

25. −5x = −3x + 6

26. 3x = 4x − 6

27. 2x − 2=4x

28. 6x − 4=2x

29. −6x +8= −2x

30. 4x − 9=3x

31. 6x = 4x − 4

32. −8x = −6x + 8

33. −8x +2= −6x + 6

34. −3x +6= −2x − 5

In Exercises 35-52, solve the equation.

35. 1 − (x − 2) = −3

36. 1 − 8(x − 8) = 17

37. −7x + 6(x + 8) = −2

38. −8x + 4(x + 7) = −12

39. 8(−6x − 1) = −8

40. −7(−2x − 4) = −14

41. −7(−4x − 6) = −14

42. −2(2x + 8) = −8

43. 2 − 9(x − 5) = −16

44. 7 − 2(x + 4) = −1

45. 7x + 2(x + 9) = −9

46. −8x + 7(x − 2) = −14

47. 2(−x + 8) = 10

48. 2(−x − 2) = 10

49. 8 + 2(x − 5) = −4

50. −5 + 2(x + 5) = −5

51. 9x − 2(x + 5) = −10

52. −8x − 5(x − 3) = 15

In Exercises 53-68, solve the equation.

53. 4(−7x + 5) + 8 = 3(−9x − 1) − 2

54. −4(−x + 9) + 5 = −(−5x − 4) − 2

55. −8(−2x − 6) = 7(5x − 1) − 2

56. 5(−4x − 8) = −9(−6x + 4) − 4

57. 2(2x − 9) + 5 = −7(−x − 8)

58. −6(−4x − 9) + 4 = −2(−9x − 8)

59. 6(−3x + 4) − 6 = −8(2x + 2) − 8

60. −5(5x − 9) − 3 = −4(2x + 5) − 6

61. 2(−2x − 3) = 3(−x + 2)

62. −2(7x + 1) = −2(3x − 7)

63. −5(−9x + 7) + 7 = −(−9x − 8)

64. 7(−2x − 6) + 1 = 9(−2x + 7)

65. 5(5x − 2) = 4(8x + 1)

66. 5(−x − 4) = −(−x + 8)

67. −7(9x − 6) = 7(5x + 7) − 7

68. −8(2x + 1) = 2(−9x + 8) − 2

1. 1

3. 4

5. −5

7. 0

9. 1

11. −2

13. −5

15. 1

17. 9

19. 7

21. −2

23. 1

25. −3

27. −1

29. 2

31. −2

33. −2

35. 6

37. 50

39. 0

41. −2

43. 7

45. −3

47. 3

49. −1

51. 0

53. 33

55. 3

57. −23

59. 21

61. −12

63. 1

65. −2

67. 0

This page titled 4.6.1: Solving Equations Involving Integers is shared under a CC BY-NC-SA license and was authored, remixed, and/or curated by David Arnold.