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

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

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Answer 1 · 2017-05-20T03:57:05

There is no solution.

Explanation:

First rearrange the equation so that a LCM of the denominator can be obtained.
$\frac{x^{2}}{x^{2} - 4} = \frac{x}{x + 2} - \frac{2}{2 - x}$ $x^2/(x^2-4)=x/(x+2)-2/(2-x)$
$\frac{x^{2}}{x^{2} - 4} = \frac{x}{x + 2} - \frac{2}{- 1 \cdot (x - 2)}$ $x^2/(x^2-4)=x/(x+2)-2/(-1*(x-2))$
$\frac{x^{2}}{x^{2} - 4} = \frac{x}{x + 2} + \frac{2}{x - 2}$ $x^2/(x^2-4)=x/(x+2)+2/(x-2)$
Now, multiply both sides by $(x - 2) (x + 2)$ $(x-2)(x+2)$ , since this is the LCM.
Doing this gets:
$x^{2} = x (x - 2) + 2 (x + 2)$ $x^2=x(x-2)+2(x+2)$
$x^{2} = x^{2} - 2 x + 2 x + 4$ $x^2=x^2-2x+2x+4$
$0 = 4$ $0=4$ which is not possible.
There is no solution.

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

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Explanation:

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