2.4 Use a General Strategy to Solve Linear 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 5

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


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 $y = - \frac{1}{3} .$

Classify Equations

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

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 $7 x plus 8$ for a different value of x, the left side will not be $−13$ .

The equation $7 x + 8 = −13$ 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 $7 x + 8 = −13$ 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.

Now let’s consider the equation $2 y + 6 = 2 (y + 3)$ . 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 $2 y$ to get the $y$ ’s to one side.
Simplify—the $y$ ’s are gone!

But $6 equals 6$ is true.

This means that the equation $2 y + 6 = 2 (y + 3)$ 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.

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


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

But $0 not equal to − 1$ .

Solving the equation $5 z = 5 z - 1$ led to the false statement $0 equals −1$ . The equation $5 z = 5 z - 1$ 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.

Example 9

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

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


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

Example 10

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

$10 + 4 (p - 5) = 0$


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 $p = \frac{5}{2} .$

Example 11

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

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


Distribute.
Combine like terms.
Subtract $14 m$ from both sides.
Simplify.
But $27 not equal to −22$ .	The equation is a contradiction. It has no solution.

**Table 2.5**
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.

Practice Makes Perfect

Solve Equations Using the General Strategy for Solving Linear Equations

In the following exercises, solve each linear equation.

$15 (y - 9) = −60$

$21 (y - 5) = −42$

$y equals 3$

$−9 (2 n + 1) = 36$

$−16 (3 n + 4) = 32$

$n equals −2$

$8 (22 + 11 r) = 0$

$5 (8 + 6 p) = 0$

$p = - \frac{4}{3}$

$− (w - 12) = 30$

$− (t - 19) = 28$

$t equals −9$

$9 (6 a + 8) + 9 = 81$

$8 (9 b - 4) - 12 = 100$

$b equals 2$

$32 + 3 (z + 4) = 41$

$21 + 2 (m - 4) = 25$

$m equals 6$

$51 + 5 (4 - q) = 56$

$−6 + 6 (5 - k) = 15$

$k = \frac{3}{2}$

$2 (9 s - 6) - 62 = 16$

$8 (6 t - 5) - 35 = −27$

$t equals 1$

$3 (10 - 2 x) + 54 = 0$

$−2 (11 - 7 x) + 54 = 4$

$x equals −2$

$\frac{2}{3} (9 c - 3) = 22$

$\frac{3}{5} (10 x - 5) = 27$

$x equals 5$

$\frac{1}{5} (15 c + 10) = c + 7$

$\frac{1}{4} (20 d + 12) = d + 7$

$d equals 1$

$18 - (9 r + 7) = −16$

$15 - (3 r + 8) = 28$

$r equals −7$

$5 - (n - 1) = 19$

$−3 - (m - 1) = 13$

$m equals −15$

$11 - 4 (y - 8) = 43$

$18 - 2 (y - 3) = 32$

$y equals −4$

$24 - 8 (3 v + 6) = 0$

$35 - 5 (2 w + 8) = −10$

$w = \frac{1}{2}$

$4 (a - 12) = 3 (a + 5)$

$−2 (a - 6) = 4 (a - 3)$

$a equals 4$

$2 (5 - u) = −3 (2 u + 6)$

$5 (8 - r) = −2 (2 r - 16)$

$r equals 8$

$3 (4 n - 1) - 2 = 8 n + 3$

$9 (2 m - 3) - 8 = 4 m + 7$

$m equals 3$

$12 + 2 (5 - 3 y) = −9 (y - 1) - 2$

$−15 + 4 (2 - 5 y) = −7 (y - 4) + 4$

$y equals −3$

$8 (x - 4) - 7 x = 14$

$5 (x - 4) - 4 x = 14$

$x equals 34$

$5 + 6 (3 s - 5) = −3 + 2 (8 s - 1)$

$−12 + 8 (x - 5) = −4 + 3 (5 x - 2)$

$x equals −6$

$4 (u - 1) - 8 = 6 (3 u - 2) - 7$

$7 (2 n - 5) = 8 (4 n - 1) - 9$

$n equals −1$

$4 (p - 4) - (p + 7) = 5 (p - 3)$

$3 (a - 2) - (a + 6) = 4 (a - 1)$

$a equals −4$

$− (9 y + 5) - (3 y - 7)$
$= 16 - (4 y - 2)$

$− (7 m + 4) - (2 m - 5)$
$= 14 - (5 m - 3)$

$m equals −4$

$4 [5 - 8 (4 c - 3)]$
$= 12 (1 - 13 c) - 8$

$5 [9 - 2 (6 d - 1)]$
$= 11 (4 - 10 d) - 139$

$d equals −3$

$3 [−9 + 8 (4 h - 3)]$
$= 2 (5 - 12 h) - 19$

$3 [−14 + 2 (15 k - 6)]$
$= 8 (3 - 5 k) - 24$

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

$5 [2 (m + 4) + 8 (m - 7)]$
$= 2 [3 (5 + m) - (21 - 3 m)]$

$10 [5 (n + 1) + 4 (n - 1)]$
$= 11 [7 (5 + n) - (25 - 3 n)]$

$n equals −5$

$5 (1.2 u - 4.8) = −12$

$4 (2.5 v - 0.6) = 7.6$

$v equals 1$

$0.25 (q - 6) = 0.1 (q + 18)$

$0.2 (p - 6) = 0.4 (p + 14)$

$p equals −34$

$0.2 (30 n + 50) = 28$

$0.5 (16 m + 34) = −15$

$m equals −4$

Classify Equations

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

$23 z + 19 = 3 (5 z - 9) + 8 z + 46$

$15 y + 32 = 2 (10 y - 7) - 5 y + 46$

identity; all real numbers

$5 (b - 9) + 4 (3 b + 9) = 6 (4 b - 5) - 7 b + 21$

$9 (a - 4) + 3 (2 a + 5) = 7 (3 a - 4) - 6 a + 7$

identity; all real numbers

$18 (5 j - 1) + 29 = 47$

$24 (3 d - 4) + 100 = 52$

conditional equation; $d = \frac{2}{3}$

$22 (3 m - 4) = 8 (2 m + 9)$

$30 (2 n - 1) = 5 (10 n + 8)$

conditional equation; $n equals 7$

$7 v + 42 = 11 (3 v + 8) - 2 (13 v - 1)$

$18 u - 51 = 9 (4 u + 5) - 6 (3 u - 10)$

contradiction; no solution

$3 (6 q - 9) + 7 (q + 4) = 5 (6 q + 8) - 5 (q + 1)$

$5 (p + 4) + 8 (2 p - 1) = 9 (3 p - 5) - 6 (p - 2)$

contradiction; no solution

$12 (6 h - 1) = 8 (8 h + 5) - 4$

$9 (4 k - 7) = 11 (3 k + 1) + 4$

conditional equation; $k equals 26$

$45 (3 y - 2) = 9 (15 y - 6)$

$60 (2 x - 1) = 15 (8 x + 5)$

contradiction; no solution

$16 (6 n + 15) = 48 (2 n + 5)$

$36 (4 m + 5) = 12 (12 m + 15)$

identity; all real numbers

$9 (14 d + 9) + 4 d = 13 (10 d + 6) + 3$

$11 (8 c + 5) - 8 c = 2 (40 c + 25) + 5$

identity; all real numbers

Everyday Math

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 $2 L + 2 (L - 2.5) = 44$ .

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 $0.05 n + 0.10 (2 n - 1) = 1.90$ .

8 nickels

Writing Exercises

Using your own words, list the steps in the general strategy for solving linear equations.

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.

Answers will vary.

What is the first step you take when solving the equation $3 - 7 (y - 4) = 38$ ? Why is this your first step?

Solve the equation $\frac{1}{4} (8 x + 20) = 3 x - 4$ explaining all the steps of your solution as in the examples in this section.

Answers will vary.

Self Check

ⓐ After completing the exercises, use this checklist to evaluate your mastery of the objective of this section.

This is a table that has three rows and four columns. In the first row, which is a header row, the cells read from left to right: “I can…,” “confidently,” “with some help,” and “no-I don’t get it!” The first column below “I can…” reads: “solve equations using the general strategy for solving linear equations,” and “classify equations.” The rest of the cells are blank.

ⓑ On a scale of 1-10, how would you rate your mastery of this section in light of your responses on the checklist? How can you improve this?