First of all, you have to determine the absolute value. Since |a|=a|a|=a if a>0a>0 and -a−a if a<0a<0, we need to determine for which xx 2x-32x−3 is greater or lesser than zero.
This is easily found:
2x-3\ge 0 \iff 2x \ge 3 \iff x \ge 3/22x−3≥0⇔2x≥3⇔x≥32
Thus, we need to study two different disequality, and then put them back together:
Case 1: x\ge 3/2x≥32
In this case, |2x-3|=2x-3|2x−3|=2x−3, and so we have
2x-3\le 4 \iff 2x \le 7 \iff x \le 7/22x−3≤4⇔2x≤7⇔x≤72
We must be very careful: our answer is accepted only if x\ge 3/2x≥32, and since we found that the answer is x \le 7/2x≤72, putting the two requests together, we have x \in [3/2, 7/2]x∈[32,72]
Case 2: x\le 3/2x≤32
In this case, |2x-3|=-2x+3|2x−3|=−2x+3, and so we have
-2x+3\le 4 \iff -2x \le 1 \iff x \ge -1/2−2x+3≤4⇔−2x≤1⇔x≥−12
As before, we must accept the request x \ge -1/2x≥−12 only for xx values smaller than 3/232, which means x \in [-1/2, 3/2]x∈[−12,32]
Our final answer is the sum of the two cases, so [-1/2, 3/2] \cup [3/2, 7/2]=[-1/2, 7/2][−12,32]∪[32,72]=[−12,72]
Here's WolframAlpha for a visual representation
