Answer
True; both the original inequality and the proposed equivalent inequality have the same solution sets.
Work Step by Step
The inequality $\frac{x-2}{x+3} \lt2$ can be re-written in the following manner: $$\frac{x-2}{x+3} \lt2$$ $$\frac{x-2}{x+3} \lt \frac{2(x+3)}{x+3}$$ $$\frac{x-2}{x+3} - \frac{2(x+3)}{x+3} \lt 0$$ $$\frac{-x -8}{x+3} \lt 0$$ $$- \frac{x + 8}{x+3} \lt 0$$ $$0 \lt \frac{x + 8}{x+3}$$ The inequality $(x-2)(x+3) \lt 2(x+3)^2$ is solved in the following manner: $$(x-2)(x+3) \lt 2(x+3)^2$$ $$x^2+x-6 \lt 2x^2 + 12x + 18$$ $$0 \lt x^2 + 11x + 24$$ $$0\lt (x+3)(x+8)$$ However, when looking at the number lines of each inequality, we realize that they both have the same solution set: $(-∞,-8)$U$(-3,∞)$. Therefore, the statement is True.
