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

Question

5858 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-09-04T12:41:00

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

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

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

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

$\frac{2}{x - 5} = \frac{- 1}{x} \leftarrow cross multiply$ $2/(x-5) = (-1)/x " "larr"cross multiply"$

$2 x = - x + 5$ $2x = -x+5$

$3 x = 5$ $3x = 5$

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

Answer 2 · 2016-09-04T13:48:46

$x = \frac{5}{3}$ $x = (5) / (3)$

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

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

$\Rightarrow 2 \cdot \frac{1}{x - 5} + \frac{1}{x} = 0$ $=> 2 cdot (1) / (x - 5) + (1) / (x) = 0$

$\Rightarrow \frac{2}{x - 5} + \frac{1}{x} = 0$ $=> (2) / (x - 5) + (1) / (x) = 0$

Let's combine the fractions:

$\Rightarrow \frac{(2) (x) + (1) (x - 5)}{(x - 5) (x)} = 0$ $=> ((2) (x) + (1) (x - 5)) / ((x - 5) (x)) = 0$

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

$\Rightarrow 3 x = 5$ $=> 3 x = 5$

$\Rightarrow x = \frac{5}{3}$ $=> x = (5) / (3)$

How do you solve 2(x-5)^-1 + 1/x = 0 2(x−5)−1+1x=02(x-5)^-1 + 1/x = 0 ?