Question

If $|2x-3|<5, x\in\mathbb{R}$, then $x$ lies in the interval

• A. [-1,1]
• B. [-1,4]
• C. (-1,4)
• D. (-2,4)
• E. [-2,4]

Hint:

Logic Tip: Pay close attention to the inequality symbols. A strict inequality (<) always translates to parentheses $()$ in interval notation, instantly eliminating options with square brackets $[]$ like A, B, and E.

Answer 1

The Correct Option is C

Concept:
The absolute value inequality $|A|<B$ (where $B>0$) implies that the expression inside the absolute value must lie strictly between $-B$ and $B$. Therefore, we can rewrite $|2x - 3|<5$ as a compound inequality: $-5<2x - 3<5$.
Step 1: Set up the compound inequality.
Remove the absolute value bars by bounding the expression: $$-5<2x - 3<5$$
Step 2: Isolate the term with x.
Add 3 to all three parts of the inequality to isolate $2x$: $$-5 + 3<2x - 3 + 3<5 + 3$$ $$-2<2x<8$$
Step 3: Solve for x.
Divide all three parts of the inequality by 2: $$\frac{-2}{2}<\frac{2x}{2}<\frac{8}{2}$$ $$-1<x<4$$
Step 4: Convert to interval notation.
Since the inequality uses strict "less than" symbols (<), the endpoints $-1$ and $4$ are not included. This corresponds to open parentheses in interval notation: $$(-1, 4)$$