Answer
$x=-\frac{7}{5}$ or $x=2$
Work Step by Step
$\displaystyle \frac{1}{x-1}+\frac{1}{x+2} = \displaystyle \frac{5}{4}$
We multiply both sides by $4(x-1)(x+2)$:
$4(x-1)(x+2)(\frac{1}{x-1}+\frac{1}{x+2}) =4(x-1)(x+2)(\displaystyle \frac{5}{4})$
$4(x+2)+4(x-1)=5(x-1)(x+2)$
Next we distribute:
$4x+8+4x-4=5x^{2}+5x-10$
And combine like terms:
$4x+8+4x-4-5x^{2}-5x+10=0$
$-5x^{2}+3x+14=0$
$5x^{2}-3x-14=0$
Finally, we factor and solve:
$(5x+7)(x-2)=0$
$5x+7=0$ or $x-2=0$
$x=-\frac{7}{5}$ or $x=2$
