Answer
$\frac{8x-3}{x(2x-1)}=\frac{3}{x}+\frac{2}{2x-1}$
Work Step by Step
$\frac{8x-3}{(2x^2-x)}$, factorising $2x^2-x=x(2x-1)$:
$\frac{8x-3}{x(2x-1)}=\frac{A}{x}+\frac{B}{2x-1}$,
$8x-3=A(2x-1)+Bx$,
$8x-3=2Ax-A+Bx$,
$8x-3=(2A+B)x-A$,
$(2A+B)x=8x$,
$2A+B=8$ and $-A=-3, A=3$:
substituting back into the equation, $6+B=8, B=2$:
thus,
$\frac{8x-3}{x(2x-1)}=\frac{3}{x}+\frac{2}{2x-1}$
