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

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

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Answer 1 · 2017-08-24T16:38:11

There are no valid solutions to this equation

Explanation:

Note that $\frac{x - 1}{x - 2} - \frac{x + 1}{x + 2} = \frac{4}{x^{2} - 4}$ $(x-1)/(x-2)-(x+1)/(x+2)=4/(x^2-4)$ is only defined if $x \neq \pm 2$ $x!=+-2$

If we attempt to solve this equation by converting all terms to the common denominator of $(x - 2) (x + 2) = x^{2} - 4$ $(x-2)(x+2)=x^2-4$
we get
$\frac{(x - 1) (x + 2)}{x^{2} - 4} - \frac{(x + 1) (x - 2)}{x^{2} - 4} = \frac{4}{x^{2} - 4}$ $((x-1)(x+2))/(x^2-4)-((x+1)(x-2))/(x^2-4)=4/(x^2-4)$

$\to (x - 1) (x + 2) - (x + 1) (x - 2) = 4$ $rarr (x-1)(x+2)-(x+1)(x-2)=4$

$\to (x^{2} + x - 2) - (x^{2} - x - 2) = 4$ $rarr (x^2+x-2)-(x^2-x-2)=4$

$rarr cancel(x^2)+xcancel(-2)cancel(-x^2)+xcancel(+2)=4$

$rarr 2x=4$

$rarr x=2$

BUT the original equation is not defined if $x=2$

Therefore there is no valid solution.

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

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Explanation:

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

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Explanation:

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