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$$x\in(3,\infty]$$Maths

Suggest Corrections

0

Similar questions
View More

People also searched for
View More