Example

Try It: Solving ∣2x−1∣≤5|2x - 1| \leq 5 and ∣4x−5∣≤3|4x - 5| \leq 3

To practice solving absolute value inequalities that use a 'less than or equal to' symbol, evaluate the expressions ∣2x−1∣≤5|2x - 1| \leq 5 and ∣4x−5∣≤3|4x - 5| \leq 3. For the first inequality, ∣2x−1∣≤5|2x - 1| \leq 5, rewrite it as the equivalent compound inequality −5≤2x−1≤5-5 \leq 2x - 1 \leq 5. Adding 11 to all parts yields −4≤2x≤6-4 \leq 2x \leq 6, and dividing by 22 gives the solution −2≤x≤3-2 \leq x \leq 3. This is graphed with closed endpoints at −2-2 and 33 and shading between them, written in interval notation as [−2,3][-2, 3]. For the second inequality, ∣4x−5∣≤3|4x - 5| \leq 3, form the compound inequality −3≤4x−5≤3-3 \leq 4x - 5 \leq 3. Adding 55 across the parts produces 2≤4x≤82 \leq 4x \leq 8, and dividing by 44 results in 12≤x≤2\frac{1}{2} \leq x \leq 2. Graphed with closed circles and shading in between, the interval notation for this solution is [12,2][\frac{1}{2}, 2].

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Updated 2026-06-03

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