Equations with x in the Denominator

Here you’ll learn how to get rid of a denominator containing variables. The good news is that the approach is the same as with getting rid of a denominator with numbers! To find the common denominator, you multiply all the different factors once. Here’s two examples:

Example 1

Solve the equation $\frac{2}{x}=3$

$\begin{array}{llll}\hfill \frac{2}{x}& =3\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill x×\frac{2}{x}& =3×x\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill 2& =3x\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill \frac{2}{3}& =\frac{\text{3}x}{\text{3}}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill \frac{2}{3}& =x\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill x& =\frac{2}{3}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\end{array}$

Example 2

Solve the equation $\frac{x+1}{x}+2=\frac{3}{x+1}+3$ for $x$

The common denominator is $x\left(x+1\right)$. Multiply both sides of the equation with the common denominator:

$\begin{array}{llll}\hfill & \phantom{=}x\left(x+1\right)×\phantom{\rule{-0.17em}{0ex}}\left(\frac{x+1}{x}+2\right)\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & =x\left(x+1\right)×\phantom{\rule{-0.17em}{0ex}}\left(\frac{3}{x+1}+3\right).\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\end{array}$

 $x\left(x+1\right)×\phantom{\rule{-0.17em}{0ex}}\left(\frac{x+1}{x}+2\right)=x\left(x+1\right)×\phantom{\rule{-0.17em}{0ex}}\left(\frac{3}{x+1}+3\right).$

This expands to
$\begin{array}{llll}\hfill & \phantom{=}x\left(x+1\right)×\frac{x+1}{x}+x\left(x+1\right)×2\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & =x\left(x+1\right)×\frac{3}{x+1}+x\left(x+1\right)×3.\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\end{array}$

 $x\left(x+1\right)×\frac{x+1}{x}+x\left(x+1\right)×2=x\left(x+1\right)×\frac{3}{x+1}+x\left(x+1\right)×3.$

Here you can cancel some factors:
$\begin{array}{llll}\hfill & \phantom{=}\text{x}\left(x+1\right)\frac{x+1}{\text{x}}+x\left(x+1\right)2\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & =x\text{(x+1)}\frac{3}{\text{x+1}}+x\left(x+1\right)3.\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\end{array}$

 $\text{x}\left(x+1\right)\frac{x+1}{\text{x}}+x\left(x+1\right)2=x\text{(x+1)}\frac{3}{\text{x+1}}+x\left(x+1\right)3.$

Now the expression simplifies to
$\begin{array}{llll}\hfill & \phantom{=}{x}^{2}+2x+1+2{x}^{2}+2x\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & =3x+3{x}^{2}+3x.\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\end{array}$

 ${x}^{2}+2x+1+2{x}^{2}+2x=3x+3{x}^{2}+3x.$

Isolate all the variables on one side and the constants on the other:
$\begin{array}{llll}\hfill -1={x}^{2}& +2{x}^{2}-3{x}^{2}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & +2x+2x-3x-3x.\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\end{array}$

 ${x}^{2}+2{x}^{2}-3{x}^{2}+2x+2x-3x-3x=-1.$

This simplifies to
 $-2x=-1.$
 $\frac{\text{−2}x}{\text{−2}}=\frac{-1}{-2}=\frac{1}{2}.$

Therefore, $x=\frac{1}{2}$.