Answer
$\frac{x}{(4x-3)(2x-1)}=\frac{3}{8x-6}+\frac{-1}{4x-2}$
Work Step by Step
$\frac{x}{(8x^2-10x+3)}$, factorising $8x^2-10x+3=(4x-3)(2x-1)$:
$\frac{x}{(4x-3)(2x-1)}=\frac{A}{4x-3}+\frac{B}{2x-1}$,
$x=A(2x-1)+B(4x-3)$,
$x=2Ax-A+4Bx-3B$,
$x=(2A+4B)x+(-A-3B)$,
$(2A+4B)x=x$,
$2A+4B=1$ and $-A-3B=0$:
multiplying by $2$ the second equation, we can solve for $B$.
$\begin{cases}
& -2A & -6B & = 0&\\
& 2A & +4B & = 1&\\
& -- & -- & -- \\
& 0 & -2B & = 1
\end{cases}$
$-2B=1, B=-\frac{1}{2}$, substituting back into the equation, $2A-2=1, A=\frac{3}{2}$
thus,
$\frac{x}{(4x-3)(2x-1)}=\frac{3}{8x-6}+\frac{-1}{4x-2}$
