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solve $-4 < 3x + 2 leq 5$. a. $x > 2$ or $x leq 1$ b. $x > -2$ or $x le…

Question

solve $-4 < 3x + 2 leq 5$.

a. $x > 2$ or $x leq 1$

b. $x > -2$ or $x leq 1$

c. $x > 2$ and $x leq 2$

d. $x > -2$ and $x leq 1$

Explanation:

Step1: Subtract 2 from all parts

$-4 - 2 < 3x + 2 - 2 \leq 5 - 2$
$\implies -6 < 3x \leq 3$

Step2: Divide all parts by 3

$\frac{-6}{3} < \frac{3x}{3} \leq \frac{3}{3}$
$\implies -2 < x \leq 1$

Step3: Match to option

This is equivalent to $x > -2$ and $x \leq 1$

Answer:

D. $x > -2$ and $x \leq 1$