2/((x^2)-x) = (1/x) + (4/((x^2)-1)

Hopefully that makes it easier to understand. The exponents are in red.

\(\displaystyle \frac{2}{x^2-x} = \frac{1}{x} + \frac{4}{x^2-1}

\)

\(\displaystyle \frac{2}{x(x-1)} = \frac{1}{x} + \frac{4}{(x+1)(x-1)}\)

common denominator is \(\displaystyle x(x+1)(x-1)\) ...

\(\displaystyle \frac{2(x+1)}{x(x-1)(x+1)} = \frac{1(x+1)(x-1)}{x(x+1)(x-1)} + \frac{4x}{x(x+1)(x-1)}\)

note that x cannot equal 0 , 1 , or -1 (why?)

with a common denominator, the numerators form the equation ...

\(\displaystyle 2(x+1) = (x+1)(x-1) + 4x\)

solve this quadratic for x