Answer
False; they have different solution sets.
Work Step by Step
$$\frac{x-2}{x+3} \lt 2$$$$x-2 \lt 2(x+3)$$ $$x-2 \lt 2x + 6$$ $$-2 - 6 \lt 2x - x$$ $$-8 \lt x$$ The exercise assumes that $x-2 \lt 2(x+3)$ is an equivalent inequality; however, it fails to take into account that the original inequality is bound by its denominator which makes it undefined at $x = -3$. Therefore, the solution for $\frac{x-2}{x+3} \lt 2$ is different to that of $x-2 \lt 2(x+3)$ which is to say $(-∞,-3)$U$(-3,∞)$ and $(-8, ∞)$ respectively.
