what happened from step 1 to step 2?
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$\displaystyle \frac{(x^2+2x-8)+(x+3)}{x^2+2x-8} = \frac{x^2+2x-8}{x^2+2x-8} + \frac{x+3}{x^2+2x-8}$
$\displaystyle \frac{x^2+2x-8+x+3}{x^2+2x-8} = \frac{x^2+2x-8}{x^2+2x-8} + \frac{x+3}{x^2+2x-8}$
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