## College Algebra 7th Edition

$x=-\frac{7}{5}$ or $x=2$
$\displaystyle \frac{1}{x-1}+\frac{1}{x+2} = \displaystyle \frac{5}{4}$ 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 $x=2$