How do you solve $\frac{2}{x + 3} - \frac{3}{4 - x} = \frac{2 x - 2}{x^{2} - x - 12}$ $2/(x+3) - 3/(4-x) =( 2x-2) /( x^2-x-12)$ ?

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Answer 1 · 2016-04-14T16:27:53

$x = - 1$ $x=-1$

$\frac{2}{x + 3} - \frac{3}{4 - x} = \frac{2 x - 2}{x^{2} - x - 12}$ $2/(x+3)-3/(4-x)=(2x-2)/(x^2-x-12)$

$\frac{2}{x + 3} + \frac{3}{x - 4} = \frac{2 x - 2}{(x - 4) (x + 3)}$ $2/(x+3)+3/(x-4)=(2x-2)/((x-4)(x+3))$

$(2(x-4)+3(x+3))/cancel(((x+3)(x-4)))=(2x-2)/cancel(((x-4)(x+3)))$

$2x-8+3x+9=2x-2$

$5x-2x=-2-1$

$3x=-3$

$x=-1$

How do you solve 2/(x+3) - 3/(4-x) =( 2x-2) /( x^2-x-12)2x+3−34−x=2x−2x2−x−122/(x+3) - 3/(4-x) =( 2x-2) /( x^2-x-12)?