Answer
$x=-\dfrac{7}{5}$ and $x=2$
Work Step by Step
$\dfrac{1}{x-1}+\dfrac{1}{x+2}=\dfrac{5}{4}$
Evaluate the sum:
$\dfrac{(x+2)+(x-1)}{(x-1)(x+2)}=\dfrac{5}{4}$
$\dfrac{2x+1}{x^{2}+x-2}=\dfrac{5}{4}$
Take $\dfrac{5}{4}$ to substract to the left side:
$\dfrac{2x+1}{x^{2}+x-2}-\dfrac{5}{4}=0$
$\dfrac{(8x+4)-(5x^{2}+5x-10)}{4x^{2}+4x-8}=0$
Take $4x^{2}+4x-8$ to multiply to the right side. We get:
$8x+4-5x^{2}-5x+10=0$
Simplify:
$-5x^{2}+3x+14=0$
Solve using the quadratic formula. Here, $a=-5$, $b=3$ and $c=14$
$x=\dfrac{-3\pm\sqrt{3^{2}-4(-5)(14)}}{2(-5)}=\dfrac{-3\pm\sqrt{9+280}}{-10}=\dfrac{-3\pm\sqrt{289}}{-10}$
$...=\dfrac{-3\pm17}{-10}$
So, our two solutions are:
$x=\dfrac{-3+17}{-10}=-\dfrac{7}{5}$ and $x=\dfrac{-3-17}{-10}=2$
