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

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Answer 1 · 2016-09-04T12:41:00

$x = 5/3$

$2(x-5)^-1 +1/x=0$

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

$x!=+5 and x!=0$

$2/(x-5) = (-1)/x " "larr"cross multiply"$

$2x = -x+5$

$3x = 5$

$x= 5/3$

Answer 2 · 2016-09-04T13:48:46

$x = (5) / (3)$

We have: $2 (x - 5)^(-1) + (1) / (x) = 0$

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

$=> 2 cdot (1) / (x - 5) + (1) / (x) = 0$

$=> (2) / (x - 5) + (1) / (x) = 0$

Let's combine the fractions:

$=> ((2) (x) + (1) (x - 5)) / ((x - 5) (x)) = 0$

$=> 2 x + x - 5 = 0$

$=> 3 x = 5$

$=> x = (5) / (3)$

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