# How do you solve 2x(x-5)^{-1}+\frac1x=0

How do you solve $2x{\left(x-5\right)}^{-1}+\frac{1}{x}=0$
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Jimmy Macias
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)
$x\ne +5$ and $x\ne 0$
$\frac{2}{x-5}=\frac{-1}{x}$
$2x=-x+5$
$3x=5$
$x=\frac{53}{}$
###### Not exactly what you’re looking for?
Jillian Edgerton
Explanation:
We have: $2{\left(x-5\right)}^{-1}+\frac{1}{x}=0$
The terms within the parentheses can be expressed as a fraction:
$⇒2\cdot \frac{1}{x-5}+\frac{1}{x}=0$
$⇒\frac{2}{x-5}+\frac{1}{x}=0$
Lets
karton