How do you solve |2x−3|≤4?

1 Answer
KillerBunny
Feb 4, 2015

First of all, you have to determine the absolute value. Since |a|=a if a>0 and −a if a<0, we need to determine for which x 2x−3 is greater or lesser than zero.

This is easily found:
2x−3≥0⇔2x≥3⇔x≥32

Thus, we need to study two different disequality, and then put them back together:

Case 1: x≥32

In this case, |2x−3|=2x−3, and so we have
2x−3≤4⇔2x≤7⇔x≤72

We must be very careful: our answer is accepted only if x≥32, and since we found that the answer is x≤72, putting the two requests together, we have x∈[32,72]

Case 2: x≤32

In this case, |2x−3|=−2x+3, and so we have
−2x+3≤4⇔−2x≤1⇔x≥−12

As before, we must accept the request x≥−12 only for x values smaller than 32, which means x∈[−12,32]

Our final answer is the sum of the two cases, so [−12,32]∪[32,72]=[−12,72]

Here's WolframAlpha for a visual representation

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