Answer
The solution is $y=-\dfrac{5}{2}$
Work Step by Step
$\dfrac{2y}{y-1}=\dfrac{5}{y}+\dfrac{10-8y}{y^{2}-y}$
Take out common factor $y$ from the denominator of the second term on the right side of the equation:
$\dfrac{2y}{y-1}=\dfrac{5}{y}+\dfrac{10-8y}{y(y-1)}$
Multiply the whole equation by $y(y-1)$
$y(y-1)\Big[\dfrac{2y}{y-1}=\dfrac{5}{y}+\dfrac{10-8y}{y(y-1)}\Big]$
$2y(y)=5(y-1)+10-8y$
$2y^{2}=5y-5+10-8y$
Take all terms to the left side of the equation and simplify:
$2y^{2}-5y+8y+5-10=0$
$2y^{2}+3y-5=0$
Solve by factoring:
$(2y+5)(y-1)=0$
Set both factors equal to $0$ and solve each individual equation for $y$:
$2y+5=0$
$2y=-5$
$y=-\dfrac{5}{2}$
$y-1=0$
$y=1$
The original equation is undefined for $y=1$, so the solution is $y=-\dfrac{5}{2}$
