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Learn to solve absolute value inequalities with step-by-step examples and exercises. Understand compound inequalities, special cases, and graphing techniques. Enhance your math skills today!
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30 Sec –5 6 5 4 –4 3 –6 –1 –2 –3 1 2 0 Which compound inequality is shown by the graph? A -3 ≥ x ≥ 1 B -3 > x > 1 C -3 < x < 1 D -3 ≤ x ≤ 1
30 Sec –5 6 5 4 –4 3 –6 –1 –2 –3 1 2 0 Which compound inequality is shown by the graph? A x ≥ 0 or x ≤ -5 B x > 0 or x ≤ -5 C x ≤ 0 or x ≥ -5 D x < 0 or x ≤ -5
Lesson2-7 Solving absolute Value inequality You'll Learn how to Solve absolute value inequality
-7 -6 -7 -6 -5 -5 -4 -4 -3 -3 -2 -2 -1 -1 0 0 1 1 2 2 3 3 4 4 5 5 6 6 7 7 2 1 EXAMPLE EXAMPLE Absolute value The absolute value of a number is the distance of that number fromzero Read as “The distance ofmy housefromthe city center is 7” | x| =7 or =7 =-7 center Solve and graph |x + 4| =2 x + 4 =2 or x + 4 =-2 x = -2 x = -6
-7 -6 -7 -6 -5 -5 -4 -4 -3 -3 -2 -1 -1 -2 0 0 1 1 2 2 3 3 4 4 5 5 6 6 7 7 3 4 EXAMPLE EXAMPLE Case1 “and” | x| <5 Read as “The distance ofmy housefromthe city center is < 5” and <5 >-5 center Solve the inequality |x| – 3<– 2 ISOLATE the absolute value expression |x| <1 Write as compound inequality and x <1 x>-1
-7 -6 -7 -6 -5 -5 -4 -4 -3 -3 -2 -1 -2 -1 0 0 1 1 2 2 3 3 4 4 5 5 6 6 7 7 5 6 EXAMPLE EXAMPLE Case2 “OR” | x| >3 Read as “The distance ofmy housefromthe city center is > 3” or >3 <-3 Solve the inequality |x| + 14≥19 ISOLATE the absolute value expression |x| ≥5 Write as compound inequality or x ≥5 x≤-5
7 EXAMPLE Special Cases: Solve and graph a) |x| + 5<2 ISOLATE the absolute value expression |x| <-3 Think: distance is never < zero Conclusion: No Solution b) |x| + 3>2 ISOLATE the absolute value expression |x| >-1 Think: distance is always > zero Conclusion: Solution is all real numbers Lesson Summary |x| <a (positive) x < a and x > -a |x| <a (negative) No Solution |x| >a (positive) x > a or x < -a |x| >a (negative) Solution is all real numbers
Exercises Solve and graph a) |x| – 3<–1 b) |x – 1| ≤2 ISOLATE |x| <2 Split x – 1 ≤2 x – 1 ≥-2 and and x<2 x>-2 d) 3 +|x + 2| >5 c) |x| + 12 >13 ISOLATE |x| >1 |x + 2| >2 Split or x<-1 x>1 or x + 2<-2 x + 2>2 e) |x + 4| – 5 >–8 f) |x + 2| + 9<7 ISOLATE |x + 2| <-2 |x + 4| >-3 Solution is all real #s No Solution