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 | |
5x - 5 < 6 | Using the addition principle for inequalities; adding 2x |
5x < 11 | Using the addition principle for inequalities; adding 5 |
x < 11/5 | Using the multiplication principle for inequalities; multiplying by or dividing by 5 |
To check, we can graph y_{1} = 3x - 5 and y_{2} = 6 - 2x. The graph shows that for x < 2.2, or x < 11/5, the graph of y_{1} lies below the graph of y_{2}, or y_{1} < y_{2}.
13 - 7x ≥ 10x - 4 | |
13 - 17x ≥ -4 | Subtracting 10x |
-17x ≥ -17 | Subtracting 13 |
x ≤ 1 | Dividing by -17 and reversing the inequality sign |
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
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 |
To check, we graph y_{1} = 2x - 5, y_{2} = -7, and y_{3} = 1. Note that for {x|x ≤ -1 or x > 3}, y_{1} ≤ y_{2} or y_{1} > y_{3}.
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.
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 |
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 |
5. State. For values of n greater than 25 hr, plan B is better for Eric.