# How do you solve 2(x-5)^-1 + 1/x = 0 ?

Sep 4, 2016

$x = \frac{5}{3}$

#### Explanation:

$2 {\left(x - 5\right)}^{-} 1 + \frac{1}{x} = 0$

Re-arrange the terms to have one fraction on each side.
(note that the negative index moves the bracket to the denominator)

$x \ne + 5 \mathmr{and} x \ne 0$

$\frac{2}{x - 5} = \frac{- 1}{x} \text{ "larr"cross multiply}$

$2 x = - x + 5$

$3 x = 5$

$x = \frac{5}{3}$

Sep 4, 2016

$x = \frac{5}{3}$

#### Explanation:

We have: $2 {\left(x - 5\right)}^{- 1} + \frac{1}{x} = 0$

The terms within the parentheses can be expressed as a fraction:

$\implies 2 \cdot \frac{1}{x - 5} + \frac{1}{x} = 0$

$\implies \frac{2}{x - 5} + \frac{1}{x} = 0$

Let's combine the fractions:

$\implies \frac{\left(2\right) \left(x\right) + \left(1\right) \left(x - 5\right)}{\left(x - 5\right) \left(x\right)} = 0$

$\implies 2 x + x - 5 = 0$

$\implies 3 x = 5$

$\implies x = \frac{5}{3}$