Question
Simplify $\frac{4 x}{x^2-1}-\frac{x+1}{x-1}$
✓
Answer
$\begin{aligned} & \frac{4 x}{(x+1)(x-1)}-\frac{x+1}{(x-1)} \\ & =\frac{4 x-(x+1)(x+1)}{(x+1)(x-1)} \\ & =\frac{4 x-\left(x^2+2 x+1\right)}{(x+1)(x-1)} \\ & =\frac{4 x-x^2-2 x-1}{(x+1)(x-1)} \\ & =\frac{-x^2+2 x-1}{(x+1)(x-1)}\end{aligned}$
$\begin{aligned} & =\frac{-\left(x^2-2 x+1\right)}{(x+1)(x-1)} \\ & =\frac{-(x-1)(x-1)}{(x+1)(x-1)} \\ & =\frac{1-x}{x+1}\end{aligned}$
$\begin{aligned} & =\frac{-\left(x^2-2 x+1\right)}{(x+1)(x-1)} \\ & =\frac{-(x-1)(x-1)}{(x+1)(x-1)} \\ & =\frac{1-x}{x+1}\end{aligned}$
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