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

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

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Answer 1 · 2017-01-17T02:23:38

${\frac{36}{7}}$ ${36/7}$

Explanation:

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

$\frac{1}{x (x - 5)} = \frac{x + 7}{x} - 1$ $1/(x(x - 5)) = (x + 7)/x - 1$

Put on a common denominator:

$\frac{1}{x (x - 5)} = \frac{(x + 7) (x - 5)}{x (x - 5)} - \frac{x (x - 5)}{x (x - 5)}$ $1/(x(x -5)) = ((x + 7)(x - 5))/(x(x - 5)) - (x(x -5))/(x(x - 5))$

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

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

$36 = 7 x$ $36 = 7x$

$x = \frac{36}{7}$ $x = 36/7$

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

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

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Explanation:

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