# Solving Linear Inequalities

An inequality is a sentence with <, >, ≤, or ≥ as its verb. An example is 3x - 5 < 6 - 2x. To solve an inequality is to find all values of the variable that make the inequality true. Each of these numbers is a solution of the inequality, and the set of all such solutions is its solution set. Inequalities that have the same solution set are called equivalent inequalities.

### Linear Inequalities

The principles for solving inequalities are similar to those for solving equations.

Principles for Solving Inequalities
For any real numbers a, b, and c :
The Addition Principle for Inequalities : If a < b is true, then a + c < b + c is true.
The Multiplication Principle for Inequalities : If a < b and c > 0 are true, then ac < bc is true. If a < b and c < 0 are true, then ac > bc is true.
Similar statements hold for a ≤ b.

When both sides of an inequality are multiplied by a negative number, we must reverse the inequality sign.
First-degree inequalities with one variable, like those in Example 1 below, are linear inequalities.

EXAMPLE 1 Solve each of the following. Then graph the solution set.
a) 3x - 5 < 6 - 2x
b) 13 - 7x ≥ 10x - 4
Solution
3x - 5 < 6 - 2x
Using the addition principle for inequalities; adding 2x
5x - 5 < 6

Using the addition principle for inequalities; adding 5
5x < 11

Using the multiplication principle for inequalities; multiplying by or dividing by 5
$x < \frac{11}{5}$

Any number less than $\frac{11}{5}$ is a solution. The solution set is $\{x|x < \frac{11}{5}\}$, or $\left(-\infin, \frac{11}{5}\right)$. The graph of the solution set is shown below. To check, we can graph y1 = 3x - 5 and y2 = 6 - 2x. The graph shows that for $x < 2.2$, or $x < \frac{11}{5}$, the graph of y1 lies below the graph of y2, or y1 < y2. Solve 13 - 7x ≥ 10x - 4
Subtracting 10x
13 - 17x ≥ -4

Subtracting 13
-17x ≥ -17

Dividing by -17 and reversing the inequality sign
x ≤ 1

The solution set is {x|x ≤ 1}, or (-∞, 1]. The graph of the solution set is shown below. ### Compound Inequalities

When two inequalities are joined by the word and or the word or, a compound inequality is formed. A compound inequality like
-3 < 2x + 5           and 2x + 5 ≤ 7
is called a conjunction, because it uses the word and. The sentence -3 < 2x + 5 ≤ 7 is an abbreviation for the preceding conjunction.
Compound inequalities can be solved using the addition and multiplication principles for inequalities.

EXAMPLE 2 Solve -3 < 2x + 5 ≤ 7. Then graph the solution set.
Solution We have

The solution set is {x| - 4 < x ≤ 1}, or (-4, 1]. The graph of the solution set is shown below. A compound inequality like 2x - 5 ≤ -7 or is called a disjunction, because it contains the word or. Unlike some conjunctions, it cannot be abbreviated; that is, it cannot be written without the word or.

EXAMPLE 3 Solve 2x - 5 ≤ -7 or 2x - 5 > 1. Then graph the solution set.
Solution We have
 -3 < 2x + 5 ≤ 7 -8 < 2x ≤ 2 Subtracting 5 -4 < x ≤ 1. Dividing by 2
 2x - 5 ≤ -7 or 2x - 5 > 1. 2x ≤ -2 or 2x > 6 Adding 5 x ≤ -1 or x > 3. Dividing by 2
The solution set is {x|x ≤ -1 or x > 3}. We can also write the solution using interval notation and the symbol for the union or inclusion of both sets: (-∞ -1] (3, ∞). The graph of the solution set is shown below. To check, we graph y1 = 2x - 5, y2 = -7, and y3 = 1. Note that for {x|x ≤ -1 or x > 3}, y1 ≤ y2 or y1 > y3. ### Inequalities with Absolute Value

Inequalities sometimes contain absolute-value notation. The following properties are used to solve them.
For a > 0 and an algebraic expression X:
|X| < a is equivalent to -a < X < a.
|X| > a is equivalent to X < -a or X > a.
Similar statements hold for |X| ≤ a and |X| ≥ a.

For example,
|x| < 3 is equivalent to -3 < x < 3;
|y| ≥ 1 is equivalent to y ≤ -1 or y ≥ 1; and |2x + 3| ≤ 4 is equivalent to -4 ≤ 2x + 3 ≤ 4.

EXAMPLE 4 Solve each of the following. Then graph the solution set.
a) |3x + 2| < 5                         b) |5 - 2x| ≥ 1
Solution
a) |3x + 2| < 5
 -5 < 3x + 2 < 5 Writing an equivalent inequality -7 < 3x < 3 Subtracting 2 -7/3 < x < 1 Dividing by 3

The solution set is {x|-7/3 < x < 1}, or (-7/3, 1). The graph of the solution set is shown below. To perform a partial check with a graphing calculator, we graph y = |3x + 2| < 5 in DOT mode. The calculator graphs a segment 1 unit above the x-axis for the values of x for which this expression for y is true. The graph shows that the solution is probably correct. b) |5 - 2x| ≥ 1

 |5 - 2x| ≤ -1 or 5 - 2x ≥ 1 Writing an equivalent inequality -2x ≤ -6 or -2x ≥ -4 Subtracting 5 x ≥ 3 or x ≤ 2 Dividing by -2 and reversing the inequality signs
The solution set is {x|x ≤ 2 or x ≥ 3}, or (-∞, 2] [3, ∞) . The graph of the solution set is shown below. ### An Application

EXAMPLE 5 Income Plans. For his house-painting job, Eric can be paid in one of two ways:
Plan A: \$250 plus \$10 per hour;
Plan B: \$20 per hour. Suppose that a job takes n hours. For what values of n is plan B better for Eric? Solution 1. Familiarize. Suppose that a job takes 20 hr. Then n = 20, and under plan A, Eric would earn \$250 + \$10.20, or \$250 + \$200, or \$450. His earnings under plan B would be \$20.20, or \$400. This shows that plan A is better for Eric if a job takes 20 hr. Similarly, if a job takes 30 hr, then n = 30, and under plan A, Eric would earn \$250 + \$10.30, or \$250 + \$300, or \$550. Under plan B, he would earn \$20.30, or \$600, so plan B is better in this case. To determine all values of n for which plan B is better for Eric, we solve an inequality. Our work in this step helps us write the inequality. 2. Translate. We translate to an inequality. 3. Carry out. We solve the inequality:  20n > 250 + 10n 10n > 250 Subtracting 10n on both sides n > 25 Dividing by 10 on both sides 4. Check. For n = 25, the income from plan A is \$250 + \$10.25, or \$250 + \$250, or \$500, and the income from plan B is \$20.25, or \$500. This shows that for a job that takes 25 hr to complete, the income is the same under either plan. In the Familiarize step, we saw that plan B pays more for a 30-hr job. Since 30 > 25, this provides a partial check of the result.We cannot check all values of n.
5. State. For values of n greater than 25 hr, plan B is better for Eric.

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