128 Chapter 2 Linear Equations and Inequalities in One Variable 2c. We must first isolate the absolute value expression and then apply property 2. u2x - 1u - 5 > 4 u2x - 1u - 5 + 5 > 4 + 5 u2x - 1u > 9 2x - 1 > 9 or 2x - 1 < -9 2x - 1 + 1 > 9 + 1 2x - 1 + 1<-9 + 1 2x > 10 2x<-8 2x 2 > 10 2 2x 2 < -8 2 x >5 x<-4 Graph: –10 –8 –6 –4 –2 0 2 4 6 8 10 Interval: (-∞, -4) ∪ (5, ∞) Student Check 2 Solve each absolute value inequality. Add 5 to each side. Simplify. Apply property 2. Add 1 to each side. Simplify. Divide each side by 2. Simplify. a. uxu > 10 b. uy - 4u ≥ 5 c. u4x + 3u - 1 > 6 Special Cases of Absolute Value Inequalities We will now investigate how to solve absolute value inequalities in which the number opposite the absolute value expression is negative. We will use the number line to solve uxu <-3 and uxu >-3. x –5 –4 –3 –2 –1 0 1 2 3 4 5 |x| 5 4 3 2 1 0 1 2 3 4 5 uxu <-3 There are no real numbers whose absolute value is less than -3. So, the solution set of this inequality is the empty set, . uxu >-3 Every real number has an absolute value that is greater than -3. So, the solution set of this inequality is all real numbers, , or (-∞, ∞). Note that the absolute value of any real number is nonnegative. That is, the absolute value of any real number is always greater than or equal to zero. This tells us two important facts: • The absolute value of a number will never be less than a negative number. • The absolute value of a number will always be greater than a negative number. These facts yield the following property. Property: Property 3 for Absolute Value Inequalities Let k be a negative real number. If uXu < k or uXu ≤ k, then the solution set is . If uXu > k or uXu ≥ k, then the solution set is (-∞, ∞) or all real numbers, . So, inequalities of the form uXu < k, uXu ≤ k, uXu > k, or uXu ≥ k, where k < 0, will either have no solution or all real numbers as solutions. Objective 3 ▶ Solve special cases of absolute value inequalities.
hendricks_intermediate_algebra_1e_ch1_3
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