Question

Solve the following inequalities.
$\frac{1}{x+2}<\frac{3}{x-3}$

Answer 1

Given that $\frac{1}{x+2}<\frac{3}{x-3}$

now we have ,

$\frac{1}{x+2}-\frac{3}{x-3}<0$

$\frac{1(x-3)-3(x+2)}{(x+2)(x-3)}<0$

$\frac{-2x-9}{(x+2)(x-3)}<0$

$\frac{2x+9}{(x+2)(x-3)}>0$

Now solve for $(2x+9)=0$ and$(x+2)(x-3)=0$

so $(2x+9)=0$

$x=\frac{-9}{2}$

and $(x+2)(x-3)=0$

$-2<x<3$

hence the final value for x for the equation $\frac{1}{x+2}<\frac{3}{x-3}$ is

$x\in (3,\infty ]$