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How do you solve $\frac{4}{x^{2} - 2 x - 3} = [- \frac{x}{x - 3}] [- \frac{1}{x + 1}]$ $4/(x^2-2x-3) = [-x/(x-3)] [-1/(x+1)]$ ?

1 Answer

Oct 23, 2015

$x = 4$ $x=4$

Explanation:

Start by simplifying the right side:
$XXX (- \frac{x}{x - 3}) (- \frac{1}{x + 1}) = \frac{x}{x^{2} - 2 x - 3}$ $color(white)("XXX")(-x/(x-3))(-1/(x+1))=x/(x^2-2x-3$

So the given equation is equivalent to
$XXX \frac{4}{x^{2} - 2 x - 3} = \frac{x}{x^{2} - 2 x - 3}$ $color(white)("XXX")4/(x^2-2x-3) = x/(x^2-2x-3)$

$\Rightarrow XXX x = 4$ $rArrcolor(white)("XXX")x=4$

This, of course, assumes $x \neq 3$ $x!=3$ and $x \neq - 1$ $x!=-1$ ; otherwise the original equation is meaningless.

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