2. Solving Linear Equations and Inequalities

2.2 Use a General Strategy to Solve Linear Equations

Learning Objectives

By the end of this section it is expected that 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

This figure is a table that has three columns and five rows. The first column is a header column, and it contains the names and numbers of each step. The second column contains further written instructions. The third column contains math. On the top row of the table, the first cell on the left reads: “Step 1. Simplify each side of the equation as much as possible.” The text in the second cell reads: “Use the Distributive Property. Notice that each side of the equation is simplified as much as possible.” The third cell contains the equation negative 6 times x plus 3, where x plus 3 is in parentheses, equals 24. Below this is the same equation with the negative 6 distributed across the parentheses: negative 6x minus 18 equals 24.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.In the fourth row of the table, the first cell says: “Step 4. Make the coefficient of the variable term equal to 1.” In the second cell, the instructions say: “Divide each side by negative 6. Simplify. The third cell contains the same equation divided by negative 6 on both sides: negative 6x over negative 6 equals 42 over negative 6, with “divided by negative 6” written in red on both sides. Below this is the answer to the equation: x equals negative 7.In the fifth row of the table, the first cell says: “Step 5. Check the solution.” In the second cell, the instructions say: “Let x equal negative 7. Simplify. Multiply.” In the third cell, there is the instruction: “Check,” and to the right of this is the original equation again: negative 6 times x plus 3, with x plus 3 in parentheses, equal 24. Below this is the same equation with negative 7 substituted in for x: negative 6 times negative 7 plus 3, with negative 7 plus 3 in parentheses, might equal 24. Below this is the equation negative 6 times negative 4 might equal 24. Below this is the equation 24 equals 24, with a check mark next to it.

TRY IT 1

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

Show answer

x=4

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 to make sure the result is a true statement.

EXAMPLE 2

Solve: \text{−}\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

Solve: \text{−}\left(y+8\right)=-2.

Show answer

y=-6

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

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

Show answer

m=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

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

Show answer

u=2

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

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

Show answer

j=\frac{5}{3}

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

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

Show answer

p=-2

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

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

Show answer

y=-\frac{17}{5}

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

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

Show answer

n=1

Classify Equations

When you solve the equation  7x+8=-13, the solution  is x=-3. This means the equation 7x+8=-13 is true when we replace the variable, x, with the value -3. We can show this by checking  the solution x=-3 and evaluating 7x+8=-13 for x=-3.

This figure shows why we can say the equation 7x plus 8 equals negative 13 is true when the variable x is replaced with the value negative 3. The first line shows the equation with negative 3 substituted in for x: 7 times negative 3 plus 8 might equal negative 13. Below this is the equation negative 21 plus 8 might equal negative 13. Below this is the equation negative 13 equals negative 13, with a check mark next to it.

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 \text{−}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

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

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

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}

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

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

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.

2.2 Exercise Set

In the following exercises, solve each linear equation.

  1. 21\left(y-5\right)=-42
  2. -16\left(3n+4\right)=32
  3. 5\left(8+6p\right)=0
  4. -\left(t-19\right)=28
  5. 21+2\left(m-4\right)=25
  6. -6+6\left(5-k\right)=15
  7. 8\left(6t-5\right)-35=-27
  8. -2\left(11-7x\right)+54=4
  9. \frac{3}{5}\left(10x-5\right)=27
  10. \frac{1}{4}\left(20d+12\right)=d+7
  11. 15-\left(3r+8\right)=28
  12. -3-\left(m-1\right)=13
  13. 18-2\left(y-3\right)=32
  14. 35-5\left(2w+8\right)=-10
  15. -2\left(a-6\right)=4\left(a-3\right)
  16. 5\left(8-r\right)=-2\left(2r-16\right)
  17. 9\left(2m-3\right)-8=4m+7
  18. -15+4\left(2-5y\right)=-7\left(y-4\right)+4
  19. 5\left(x-4\right)-4x=14
  20. -12+8\left(x-5\right)=-4+3\left(5x-2\right)
  21. 7\left(2n-5\right)=8\left(4n-1\right)-9
  22. 3\left(a-2\right)-\left(a+6\right)=4\left(a-1\right)
  23. -\left(7m+4\right)-\left(2m-5\right) =14-\left(5m-3\right)
  24. 5\left[9-2\left(6d-1\right)\right] =11\left(4-10d\right)-139
  25. 3\left[-14+2\left(15k-6\right)\right] =8\left(3-5k\right)-24
  26. 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]
  27. 4\left(2.5v-0.6\right)=7.6
  28. 0.2\left(p-6\right)=0.4\left(p+14\right)
  29. 0.5\left(16m+34\right)=-15

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

  1. 15y+32=2\left(10y-7\right)-5y+46
  2. 9\left(a-4\right)+3\left(2a+5\right)=7\left(3a-4\right)-6a+7
  3. 24\left(3d-4\right)+100=52
  4. 30\left(2n-1\right)=5\left(10n+8\right)
  5. 18u-51=9\left(4u+5\right)-6\left(3u-10\right)
  6. 5\left(p+4\right)+8\left(2p-1\right)=9\left(3p-5\right)-6\left(p-2\right)
  7. 9\left(4k-7\right)=11\left(3k+1\right)+4
  8. 60\left(2x-1\right)=15\left(8x+5\right)
  9. 36\left(4m+5\right)=12\left(12m+15\right)
  10. 11\left(8c+5\right)-8c=2\left(40c+25\right)+5
  11. Coins. Marta 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.

Answers

  1. y=3
  2. n=-2
  3. p=-\frac{4}{3}
  4. t=47
  5. m=6
  6. k=\frac{3}{2}
  7. t=1
  8. x=-2
  9. x=5
  10. d=1
  11. r=-7
  12. m=-15
  13. y=-4
  14. w=\frac{1}{2}
  15. a=4
  16. r=8
  17. m=3
  18. y=-3
  19. x=34
  20. x=-6
  21. n=-1
  22. a=-4
  23. m=\frac{4}{5}
  24. d=-3
  25. k=\frac{3}{5}
  26. n=-5
  27. v=1
  28. p=-34
  29. m=-4
  30. identity; all real numbers
  31. identity; all real numbers
  32. conditional equation; d=\frac{2}{3}
  33. conditional equation; n=7
  34. contradiction; no solution
  35. contradiction; no solution
  36. conditional equation; k=26
  37. contradiction; no solution
  38. identity; all real numbers
  39. identity; all real numbers
  40. 8 nickels

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 Adaptation Statement for more information.

License

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Business/Technical Mathematics Copyright © 2021 by Izabela Mazur and Kim Moshenko is licensed under a Creative Commons Attribution 4.0 International License, except where otherwise noted.

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