Chapter 1.5: Use a General Strategy to Solve Linear Equations

Mike LePine

Chapter 1.5: Use a General Strategy to Solve Linear Equations

Learning Objectives

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

Solve equations using a general strategy
Classify equations

Solve Equations Using the General Strategy

Until now we have dealt with solving one specific form of a linear equation. It is time now to lay out one overall strategy that can be used to solve any linear equation. Some equations we solve will not require all these steps to solve, but many will.

Beginning by simplifying each side of the equation makes the remaining steps easier.

EXAMPLE 1. How to Solve Linear Equations Using the General Strategy

Solve: $-6\left(x+3\right)=24$ .

Solution

In the second row of the table, the first cell says: “Step 2. Collect all variable terms on one side of the equation.” In the second cell, the instructions say: “Nothing to do—all x’s are on the left side. The third cell is blank. In the third row of the table, the first cell says: “Step 3. Collect constant terms on the other side of the equation. In the second cell, the instructions say: “To get constants only on the right, add 18 to each side. Simplify.” The third cell contains the same equation with 18 added to both sides: negative 6x minus 18 plus 18 equals 24 plus 18. Below this is the equation negative 6x equals 42.

TRY IT 1.1

Solve: $5\left(x+3\right)=35$ .

Show answer

$x=4$

TRY IT 1.2

Solve: $6\left(y-4\right)=-18$ .

Show answer

$y=1$

General strategy for solving linear equations.

Simplify each side of the equation as much as possible.
Use the Distributive Property to remove any parentheses.
Combine like terms.
Collect all the variable terms on one side of the equation.
Use the Addition or Subtraction Property of Equality.
Collect all the constant terms on the other side of the equation.
Use the Addition or Subtraction Property of Equality.
Make the coefficient of the variable term to equal to 1.
Use the Multiplication or Division Property of Equality.
State the solution to the equation.
Check the solution. Substitute the solution into the original equation to make sure the result is a true statement.

EXAMPLE 2

Solve: $-\left(y+9\right)=8$ .

Solution


Simplify each side of the equation as much as possible by distributing.
The only $y$ term is on the left side, so all variable terms are on the left side of the equation.
Add $9$ to both sides to get all constant terms on the right side of the equation.
Simplify.
Rewrite $-y$ as $-1y$ .
Make the coefficient of the variable term to equal to $1$ by dividing both sides by $-1$ .
Simplify.
Check: Let $y=-17$ .

TRY IT 2.1

Solve: $-\left(y+8\right)=-2$ .

Show answer

$y=-6$

TRY IT 2.2

Solve: $-\left(z+4\right)=-12$ .

Show answer

$z=8$

EXAMPLE 3

Solve: $5\left(a-3\right)+5=-10$ .

Solution


Simplify each side of the equation as much as possible.
Distribute.
Combine like terms.
The only $a$ term is on the left side, so all variable terms are on one side of the equation.
Add $10$ to both sides to get all constant terms on the other side of the equation.
Simplify.
Make the coefficient of the variable term to equal to $1$ by dividing both sides by $5$ .
Simplify.
Check:
Let $a=0$ .

TRY IT 3.1

Solve: $2\left(m-4\right)+3=-1$ .

Show answer

$m=2$

TRY IT 3.2

Solve: $7\left(n-3\right)-8=-15$ .

Show answer

$n=2$

EXAMPLE 4

Solve: $\frac{2}{3}\left(6m-3\right)=8-m$ .

Solution


Distribute.
Add $m$ to get the variables only to the left.
Simplify.
Add $2$ to get constants only on the right.
Simplify.
Divide by $5$ .
Simplify.
Check:
Let $m=2$ .

TRY IT 4.1

Solve: $\frac{1}{3}\left(6u+3\right)=7-u$ .

Show answer

$u=2$

TRY IT 4.2

Solve: $\frac{2}{3}\left(9x-12\right)=8+2x$ .

Show answer

$x=4$

EXAMPLE 5

Solve: $8-2\left(3y+5\right)=0$ .

Solution


Simplify—use the Distributive Property.
Combine like terms.
Add $2$ to both sides to collect constants on the right.
Simplify.
Divide both sides by $-6$ .
Simplify.
Check: Let $y=-\frac{1}{3}$ .

TRY IT 5.1

Solve: $12-3\left(4j+3\right)=-17$ .

Show answer

$j=\frac{5}{3}$

TRY IT 5.2

Solve: $-6-8\left(k-2\right)=-10$ .

Show answer

$k=\frac{5}{2}$

EXAMPLE 6

Solve: $4\left(x-1\right)-2=5\left(2x+3\right)+6$ .

Solution


Distribute.
Combine like terms.
Subtract $4x$ to get the variables only on the right side since $10$ > $4$ .
Simplify.
Subtract $21$ to get the constants on left.
Simplify.
Divide by 6.
Simplify.
Check:
Let $x=-\frac{9}{2}$ .

TRY IT 6.1

Solve: $6\left(p-3\right)-7=5\left(4p+3\right)-12$ .

Show answer

$p=-2$

TRY IT 6.2

Solve: $8\left(q+1\right)-5=3\left(2q-4\right)-1$ .

Show answer

$q=-8$

EXAMPLE 7

Solve: $10\left[3-8\left(2s-5\right)\right]=15\left(40-5s\right)$ .

Solution


Simplify from the innermost parentheses first.
Combine like terms in the brackets.
Distribute.
Add $160s$ to get the s’s to the right.
Simplify.
Subtract 600 to get the constants to the left.
Simplify.
Divide.
Simplify.
Check:
Substitute $s=-2$ .

TRY IT 7.1

Solve: $6\left[4-2\left(7y-1\right)\right]=8\left(13-8y\right)$ .

Show answer

$y=-\frac{17}{5}$

TRY IT 7.2

Solve: $12\left[1-5\left(4z-1\right)\right]=3\left(24+11z\right)$ .

Show answer

$z=0$

EXAMPLE 8

Solve: $0.36\left(100n+5\right)=0.6\left(30n+15\right)$ .

Solution


Distribute.
Subtract $18n$ to get the variables to the left.
Simplify.
Subtract $1.8$ to get the constants to the right.
Simplify.
Divide.
Simplify.
Check:
Let $n=0.4$ .

TRY IT 8.1

Solve: $0.55\left(100n+8\right)=0.6\left(85n+14\right)$ .

Show answer

$n=1$

TRY IT 8.2

Solve: $0.15\left(40m-120\right)=0.5\left(60m+12\right)$ .

Show answer

$m=-1$

Classify Equations

Consider the equation we solved at the start of the last section, $7x+8=-13$ . The solution we found was $x=-3$ . This means the equation $7x+8=-13$ is true when we replace the variable, x, with the value $-3$ . We showed this when we checked the solution $x=-3$ and evaluated $7x+8=-13$ for $x=-3$ .

If we evaluate $7x+8$ for a different value of x, the left side will not be $-13$ .

The equation $7x+8=-13$ is true when we replace the variable, x, with the value $-3$ , but not true when we replace x with any other value. Whether or not the equation $7x+8=-13$ is true depends on the value of the variable. Equations like this are called conditional equations.

All the equations we have solved so far are conditional equations.

Conditional equation

An equation that is true for one or more values of the variable and false for all other values of the variable is a conditional equation.

Now let’s consider the equation $2y+6=2\left(y+3\right)$ . Do you recognize that the left side and the right side are equivalent? Let’s see what happens when we solve for y.


Distribute.
Subtract $2y$ to get the $y$ ’s to one side.
Simplify—the $y$ ’s are gone!

But $6=6$ is true.

This means that the equation $2y+6=2\left(y+3\right)$ is true for any value of y. We say the solution to the equation is all of the real numbers. An equation that is true for any value of the variable like this is called an identity.

Identity

An equation that is true for any value of the variable is called an identity.

The solution of an identity is every real number.

What happens when we solve the equation $5z=5z-1$ ?


Subtract $5z$ to get the constant alone on the right.
Simplify—the $z$ ’s are gone!

But $0\ne 1$ .

Solving the equation $5z=5z-1$ led to the false statement $0=-1$ . The equation $5z=5z-1$ will not be true for any value of z. It has no solution. An equation that has no solution, or that is false for all values of the variable, is called a contradiction.

Contradiction

An equation that is false for all values of the variable is called a contradiction.

A contradiction has no solution.

EXAMPLE 9

Classify the equation as a conditional equation, an identity, or a contradiction. Then state the solution.

$6\left(2n-1\right)+3=2n-8+5\left(2n+1\right)$

Solution


Distribute.
Combine like terms.
Subtract $12n$ to get the $n$ ’s to one side.
Simplify.
This is a true statement.	The equation is an identity. The solution is every real number.

TRY IT 9.1

Classify the equation as a conditional equation, an identity, or a contradiction and then state the solution:

$4+9\left(3x-7\right)=-42x-13+23\left(3x-2\right)$

Show answer

identity; all real numbers

TRY IT 9.2

Classify the equation as a conditional equation, an identity, or a contradiction and then state the solution:

$8\left(1-3x\right)+15\left(2x+7\right)=2\left(x+50\right)+4\left(x+3\right)+1$

Show answer

identity; all real numbers

EXAMPLE 10

Classify as a conditional equation, an identity, or a contradiction. Then state the solution.

$10+4\left(p-5\right)=0$

Solution


Distribute.
Combine like terms.
Add $10$ to both sides.
Simplify.
Divide.
Simplify.
The equation is true when $p=\frac{5}{2}$ .	This is a conditional equation. The solution is $p=\frac{5}{2}$ .

TRY IT 10.1

Classify the equation as a conditional equation, an identity, or a contradiction and then state the solution: $11\left(q+3\right)-5=19$

Show answer

conditional equation; $q=\frac{9}{11}$

TRY IT 10.2

Classify the equation as a conditional equation, an identity, or a contradiction and then state the solution: $6+14\left(k-8\right)=95$

Show answer

conditional equation; $k=\frac{193}{14}$

EXAMPLE 11

Classify the equation as a conditional equation, an identity, or a contradiction. Then state the solution.

$5m+3\left(9+3m\right)=2\left(7m-11\right)$

Solution


Distribute.
Combine like terms.
Subtract $14m$ from both sides.
Simplify.
But $27\ne -22$ .	The equation is a contradiction. It has no solution.

TRY IT 11.1

Classify the equation as a conditional equation, an identity, or a contradiction and then state the solution:

$12c+5\left(5+3c\right)=3\left(9c-4\right)$

Show answer

contradiction; no solution

TRY IT 11.2

Classify the equation as a conditional equation, an identity, or a contradiction and then state the solution:

$4\left(7d+18\right)=13\left(3d-2\right)-11d$

Show answer

contradiction; no solution

Type of equation – Solution

Type of equation	What happens when you solve it?	Solution
Conditional Equation	True for one or more values of the variables and false for all other values	One or more values
Identity	True for any value of the variable	All real numbers
Contradiction	False for all values of the variable	No solution

Key Concepts

General Strategy for Solving Linear Equations
1. Simplify each side of the equation as much as possible.
  Use the Distributive Property to remove any parentheses.
  Combine like terms.
2. Collect all the variable terms on one side of the equation.
  Use the Addition or Subtraction Property of Equality.
3. Collect all the constant terms on the other side of the equation.
  Use the Addition or Subtraction Property of Equality.
4. Make the coefficient of the variable term to equal to 1.
  Use the Multiplication or Division Property of Equality.
  State the solution to the equation.
5. Check the solution.
  Substitute the solution into the original equation.

Glossary

conditional equation: An equation that is true for one or more values of the variable and false for all other values of the variable is a conditional equation.

contradiction: An equation that is false for all values of the variable is called a contradiction. A contradiction has no solution.

identity: An equation that is true for any value of the variable is called an identity. The solution of an identity is all real numbers.

Practice Makes Perfect

Solve Equations Using the General Strategy for Solving Linear Equations

In the following exercises, solve each linear equation.

1. $21\left(y-5\right)=-42$	2. $15\left(y-9\right)=-60$
3. $-16\left(3n+4\right)=32$	4. $-9\left(2n+1\right)=36$
5. $5\left(8+6p\right)=0$	6. $8\left(22+11r\right)=0$
7. $-\left(t-19\right)=28$	8. $-\left(w-12\right)=30$
9. $21+2\left(m-4\right)=25$	10. $32+3\left(z+4\right)=41$
11. $-6+6\left(5-k\right)=15$	12. $51+5\left(4-q\right)=56$
13. $8\left(6t-5\right)-35=-27$	14. $2\left(9s-6\right)-62=16$
15. $-2\left(11-7x\right)+54=4$	16. $3\left(10-2x\right)+54=0$
17. $\frac{3}{5}\left(10x-5\right)=27$	18. $\frac{2}{3}\left(9c-3\right)=22$
19. $\frac{1}{4}\left(20d+12\right)=d+7$	20. $\frac{1}{5}\left(15c+10\right)=c+7$
21. $15-\left(3r+8\right)=28$	22. $18-\left(9r+7\right)=-16$
23. $-3-\left(m-1\right)=13$	24. $5-\left(n-1\right)=19$
25. $18-2\left(y-3\right)=32$	26. $11-4\left(y-8\right)=43$
27. $35-5\left(2w+8\right)=-10$	28. $24-8\left(3v+6\right)=0$
29. $-2\left(a-6\right)=4\left(a-3\right)$	30. $4\left(a-12\right)=3\left(a+5\right)$
31. $5\left(8-r\right)=-2\left(2r-16\right)$	32. $2\left(5-u\right)=-3\left(2u+6\right)$
33. $9\left(2m-3\right)-8=4m+7$	34. $3\left(4n-1\right)-2=8n+3$
35. $-15+4\left(2-5y\right)=-7\left(y-4\right)+4$	36. $12+2\left(5-3y\right)=-9\left(y-1\right)-2$
37. $5\left(x-4\right)-4x=14$	38. $8\left(x-4\right)-7x=14$
39. $-12+8\left(x-5\right)=-4+3\left(5x-2\right)$	40. $5+6\left(3s-5\right)=-3+2\left(8s-1\right)$
41. $7\left(2n-5\right)=8\left(4n-1\right)-9$	42. $4\left(u-1\right)-8=6\left(3u-2\right)-7$
43. $3\left(a-2\right)-\left(a+6\right)=4\left(a-1\right)$	44. $4\left(p-4\right)-\left(p+7\right)=5\left(p-3\right)$
45. $-\left(7m+4\right)-\left(2m-5\right)$ $=14-\left(5m-3\right)$	46. $-\left(9y+5\right)-\left(3y-7\right)$ $=16-\left(4y-2\right)$
47. $5\left[9-2\left(6d-1\right)\right]$ $=11\left(4-10d\right)-139$	48. $4\left[5-8\left(4c-3\right)\right]$ $=12\left(1-13c\right)-8$
49. $3\left[-14+2\left(15k-6\right)\right]$ $=8\left(3-5k\right)-24$	50. $3\left[-9+8\left(4h-3\right)\right]$ $=2\left(5-12h\right)-19$
51. $10\left[5\left(n+1\right)+4\left(n-1\right)\right]$ $=11\left[7\left(5+n\right)-\left(25-3n\right)\right]$	52. $5\left[2\left(m+4\right)+8\left(m-7\right)\right]$ $=2\left[3\left(5+m\right)-\left(21-3m\right)\right]$
53. $4\left(2.5v-0.6\right)=7.6$	54. $5\left(1.2u-4.8\right)=-12$
55. $0.2\left(p-6\right)=0.4\left(p+14\right)$	56. $0.25\left(q-6\right)=0.1\left(q+18\right)$
57. $0.5\left(16m+34\right)=-15$	58. $0.2\left(30n+50\right)=28$

Classify Equations

In the following exercises, classify each equation as a conditional equation, an identity, or a contradiction and then state the solution.

59. $15y+32=2\left(10y-7\right)-5y+46$	60. $23z+19=3\left(5z-9\right)+8z+46$
61. $9\left(a-4\right)+3\left(2a+5\right)=7\left(3a-4\right)-6a+7$	62. $5\left(b-9\right)+4\left(3b+9\right)=6\left(4b-5\right)-7b+21$
63. $24\left(3d-4\right)+100=52$	64. $18\left(5j-1\right)+29=47$
65. $30\left(2n-1\right)=5\left(10n+8\right)$	66. $22\left(3m-4\right)=8\left(2m+9\right)$
67. $18u-51=9\left(4u+5\right)-6\left(3u-10\right)$	68. $7v+42=11\left(3v+8\right)-2\left(13v-1\right)$
69. $5\left(p+4\right)+8\left(2p-1\right)=9\left(3p-5\right)-6\left(p-2\right)$	70. $3\left(6q-9\right)+7\left(q+4\right)=5\left(6q+8\right)-5\left(q+1\right)$
71. $9\left(4k-7\right)=11\left(3k+1\right)+4$	72. $12\left(6h-1\right)=8\left(8h+5\right)-4$
73. $60\left(2x-1\right)=15\left(8x+5\right)$	74. $45\left(3y-2\right)=9\left(15y-6\right)$
75. $36\left(4m+5\right)=12\left(12m+15\right)$	76. $16 \left(6n+15 \right)= 48 \left(2n+5\right)$
77. $11\left(8c+5\right)-8c=2\left(40c+25\right)+5$	78. $9\left(14d+9\right)+4d=13\left(10d+6\right)+3$

Everyday Math

79. Coins. Rhonda has $1.90 in nickels and dimes. The number of dimes is one less than twice the number of nickels. Find the number of nickels, n, by solving the equation $0.05n+0.10\left(2n-1\right)=1.90$ .

80. Fencing. Micah has 44 feet of fencing to make a dog run in his yard. He wants the length to be 2.5 feet more than the width. Find the length, L, by solving the equation $2L+2\left(L-2.5\right)=44$ .

Writing Exercises

81. Explain why you should simplify both sides of an equation as much as possible before collecting the variable terms to one side and the constant terms to the other side.	82. Using your own words, list the steps in the general strategy for solving linear equations.
83. Solve the equation $\frac{1}{4}\left(8x+20\right)=3x-4$ explaining all the steps of your solution as in the examples in this section.	84. What is the first step you take when solving the equation $3-7\left(y-4\right)=38$ ? Why is this your first step?

Answers

1. $y=3$	3. $n=-2$	5. $p=-\frac{4}{3}$
7. $t=-9$	9. $m=6$	11. $k=\frac{3}{2}$
13. $t=1$	15. $x=-2$	17. $x=5$
19. $d=1$	21. $r=-7$	23. $m=-15$
25. $y=-4$	27. $w=\frac{1}{2}$	29. $a=4$
31. $r=8$	33. $m=3$	35. $y=-3$
37. $x=34$	39. $x=-6$	41. $n=-1$
43. $a=-4$	45. $m=-4$	47. $d=-3$
49. $k=\frac{3}{5}$	51. $n=-5$	53. $v=1$
55. $p=-34$	57. $m=-4$	59. identity; all real numbers
61. identity; all real numbers	63. conditional equation; $d=\frac{2}{3}$	65. conditional equation; $n=7$
67. contradiction; no solution	69. contradiction; no solution	71. conditional equation; $k=26$
73. contradiction; no solution	75. identity; all real numbers	77. identity; all real numbers
79. 8 nickels	81. Answers will vary.	83. Answers will vary.

Attributions

This chapter has been adapted from “Use a General Strategy to Solve Linear Equations” in Elementary Algebra (OpenStax) by Lynn Marecek and MaryAnne Anthony-Smith, which is under a CC BY 4.0 Licence. Adapted by Izabela Mazur. See the Copyright page for more information.

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Solve Equations Using the General Strategy

Classify Equations

Key Concepts

Glossary

Practice Makes Perfect

Solve Equations Using the General Strategy for Solving Linear Equations

Classify Equations

Everyday Math

Writing Exercises

Answers

Attributions

License

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