How do you solve $\frac{x + 1}{x - 3} = 4 - \frac{12}{x^{2} - 2 x - 3}$ $(x+1)/(x-3) = 4 - 12/(x^2-2x-3)$ ?

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How do you solve $\frac{x + 1}{x - 3} = 4 - \frac{12}{x^{2} - 2 x - 3}$ $(x+1)/(x-3) = 4 - 12/(x^2-2x-3)$ ?

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Answer 1 · 2015-05-31T12:54:30

Given $\frac{x + 1}{x - 3} = 4 - \frac{12}{x^{2} - 2 x - 3}$ $(x+1)/(x-3) = 4 - 12/(x^2-2x-3)$
and
assuming that $x^{2} - 2 x - 3 \neq 0$ $x^2-2x-3 != 0$ (otherwise we have an undefined division)

Note that $x^{2} - 2 x - 3 = (x - 3) ((x + 1)$ $x^2-2x-3= (x-3)((x+1)$

So we can re-write the equation with a common denominators as
$XXXXX$ $color(white)("XXXXX")$ $\frac{(x + 1) (x + 1)}{x^{2} - 2 x - 3} = \frac{4 (x^{2} - 2 x - 3) - 12}{x^{2} - 2 x - 3}$ $((x+1)(x+1))/(x^2-2x-3) = (4(x^2-2x-3) - 12)/(x^2-2x-3)$
and since $x^{2} - 2 x - 3 \neq 0$ $x^2-2x-3 != 0$
$XXXXX$ $color(white)("XXXXX")$ $((x + 1) (x + 1)) = (4 (x^{2} - 2 x - 3) - 12)$ $((x+1)(x+1)) = (4(x^2-2x-3) - 12)$

Simplifying
$XXXXX$ $color(white)("XXXXX")$ $x^{2} + 2 x + 1 = 4 x^{2} - 8 x - 24$ $x^2+2x+1 = 4x^2-8x -24$

$XXXXX$ $color(white)("XXXXX")$ $3 x^{2} - 10 x - 25 = 0$ $3x^2-10x-25 = 0$

$XXXXX$ $color(white)("XXXXX")$ $(3 x + 5) (x - 5) = 0$ $(3x+5)(x-5) = 0$

$x = - \frac{5}{3}$ $x= -5/3$ or $x = 5$ $x=5$

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How do you solve $\frac{x + 1}{x - 3} = 4 - \frac{12}{x^{2} - 2 x - 3}$ $(x+1)/(x-3) = 4 - 12/(x^2-2x-3)$ ?

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How do you solve x+1x−3=4−12x2−2x−3(x+1)/(x-3) = 4 - 12/(x^2-2x-3)?

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How do you solve $\frac{x + 1}{x - 3} = 4 - \frac{12}{x^{2} - 2 x - 3}$ $(x+1)/(x-3) = 4 - 12/(x^2-2x-3)$ ?