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Use the definition of absolute value inequalities: ∣u∣0k>0. Let u=2x−3u = 2x - 3 and k=5k=5. Then

−5<2x−3<5-5 < 2x - 3 < 5.

Add 3 throughout: −2<2x<8-2 < 2x < 8.

Divide by 2: −1

Thus the interval is (−1,4)(-1, 4), choice A. Pitfalls include reversing inequality directions (unnecessary here since we divided by a positive) or forgetting to apply the transformation to both sides when adding or dividing. Some test takers instead split cases: 2x−3<52x - 3 < 5 and 2x−3>−52x - 3 > -5, which yields the same result. A quick check: pick a value inside (e.g., x=0x=0): ∣−3∣=3<5|-3|=3<5 → true. Pick a boundary value x=−1x=-1: ∣2(−1)−3∣=∣−5∣=5|2(-1)-3|=| -5|=5 → not allowed (strict inequality), consistent with open interval endpoints. GRE inequality items reward systematic algebra and attention to strict vs. non-strict symbols; draw a number line if it helps visualize the solution.