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

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

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Answer 1 · 2015-10-05T21:30:15

$x = - 1$ $x=-1$ and $x = - 10$ $x=-10$

Explanation:

We have, at first, $\frac{x}{x + 2} - \frac{x + 2}{x - 2} = \frac{x + 3}{x - 2}$ $(x)/(x+2) - (x+2)/(x-2) = (x+3)/(x-2)$ .
We pass $- \frac{x + 2}{x - 2}$ $-(x+2)/(x-2)$ to the other side adding.
We now have $\frac{x}{x + 2} = \frac{x + 3}{x - 2} + \frac{x + 2}{x - 2}$ $(x)/(x+2) = (x+3)/(x-2) + (x+2)/(x-2)$ .
As the denominator is common on both fractions of the right side, we add them.
We get $\frac{x}{x + 2} = \frac{2 x + 5}{x - 2}$ $(x)/(x+2) = (2x + 5)/(x-2)$ .
We now cross multiply to get $x (x - 2) = (x + 2) (2 x + 5)$ $x(x-2) = (x+2)(2x+5)$ .
We multiply and get $x^{2} - 2 x = 2 x^{2} + 9 x + 10$ $x^2 - 2x = 2x^2 + 9x + 10$ .
We pass the left hand side working to the right, getting $x^{2} + 11 x + 10 = 0$ $x^2 + 11x + 10 = 0$ as an answer.
We factorise to get $(x + 1) (x + 10) = 0$ $(x+1)(x+10) = 0$ .
Therefore, the answers are $x = - 1$ $x = -1$ and $x = - 10$ $x = -10$ .
Hope it helps :D .

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

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Explanation:

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