How do you solve x/(x^2-8)=2/x?

Jul 27, 2016

$x = \pm 4$

Explanation:

First, move everything from the denominator to the numerator
we do this by multiplying by the LCM of denominators on each side.

$x \left({x}^{2} - 8\right) \cdot \frac{x}{{x}^{2} - 8} = \frac{2}{x} \cdot x \left({x}^{2} - 8\right)$

the $\left({x}^{2} - 8\right)$ on the left side cancels out and the $x$ on the right side cancels out

$x \left(\cancel{{x}^{2} - 8}\right) \cdot \frac{x}{\cancel{{x}^{2} - 8}} = \frac{2}{\cancel{x}} \cdot \cancel{x} \left({x}^{2} - 8\right)$

which leaves us with:

$x \cdot x = 2 \cdot \left({x}^{2} - 8\right)$

after this we now have
${x}^{2} = 2 \left({x}^{2} - 8\right)$

next we remove the parentheses by multiplying each term by 2
now we have
${x}^{2} = 2 {x}^{2} - 16$

next we will move the 16 to the other side to avoid working with negative numbers
$16 + {x}^{2} = 2 {x}^{2}$

then we will combine like terms by subtracting ${x}^{2}$
$16 = 2 {x}^{2} - {x}^{2}$ leaving us with $16 = {x}^{2}$

then we will get rid of the ${x}^{2}$ by taking the square root of both sides
$\pm \sqrt{16}$=${\sqrt{x}}^{2}$

now we have our final answer of
$x = \pm 4$