How do you solve $1/(x^2-5x)=(x+7)/x-1$ and check for extraneous solutions?

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Answer 1 · 2017-01-17T02:23:38

${36/7}$

Recall that $x^2 - 5x$ can be written as $x(x - 5)$ .

$1/(x(x - 5)) = (x + 7)/x - 1$

Put on a common denominator:

$1/(x(x -5)) = ((x + 7)(x - 5))/(x(x - 5)) - (x(x -5))/(x(x - 5))$

$1 = x^2 +7x - 5x - 35 - (x^2 - 5x)$

$1 = x^2 + 7x - 5x - 35 - x^2 + 5x$

$36 = 7x$

$x = 36/7$

This is not extraneous, as $x != 0, 5$ .

Hopefully this helps!

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