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Question

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


Solution

Given that $$\dfrac {1}{x+2}<\dfrac {3}{x-3}$$
now we have ,
$$\dfrac {1}{x+2}-\dfrac {3}{x-3}<0$$
$$\dfrac {1(x-3)-3(x+2)}{(x+2)(x-3)}<0$$
$$\dfrac {-2x-9}{(x+2)(x-3)}<0$$
$$\dfrac {2x+9}{(x+2)(x-3)}>0$$
Now solve for  $$(2x+9)=0$$ and$$(x+2)(x-3)=0$$
so $$(2x+9)=0$$
$$x=\dfrac {-9}{2}$$
and $$(x+2)(x-3)=0$$
$$-2<x<3$$
hence the final value for x for the equation $$\dfrac {1}{x+2}<\dfrac {3}{x-3}$$ is
$$x\in(3,\infty]$$


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