Answer
$-\frac{x-1}{x+1}$
Work Step by Step
$=\frac{x^2+2x-3}{x^2-2x-3}\times\frac{3-x}{3+x}$
Simplify the fraction $\frac{x^2+2x-3}{x^2-2x-3}$:
$=\frac{(x+3)(x-1)}{(x+1)(x-3)}$
Multiply the fractions:
$=\frac{(x+3)(x-1)}{(x+1)(x-3)}\times\frac{3-x}{3+x}$
$=\frac{(x+3)(x-1)(-x+3)}{(x+1)(x-3)(x+3)}$
Extract the negative sign from $-x+3$:
$=\frac{-(x-3)(x-1)(x+3)}{(x+1)(x-3)(x+3)}$
Simplify the fraction:
$=-\frac{x-1}{x+1}$
