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Equations with x in the Denominator

Having learned how to use common denominators to get rid of fractions in equations, you might be wondering how to isolate a variable that is in the denominator. This is a great question!

The good news is that the approach is the same as with getting rid of a denominator that contains only 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}{l} \frac{2}{x} & = 3 \\ x \cdot \frac{2}{x} & = 3 \cdot x \\ 2 & = 3 x \\ \frac{2}{3} & = \frac{3 x}{3} \\ \frac{2}{3} & = x \\ x & = \frac{2}{3} \end{array}

Example 2

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

The common denominator is $x (x + 1)$ . Multiply both sides of the equation by the common denominator:

\begin{array}{l} x (x + 1) (\frac{x + 1}{x} + 2) \\ = x (x + 1) (\frac{3}{x + 1} + 3) \end{array}

x (x + 1) \cdot (\frac{x + 1}{x} + 2) = x (x + 1) \cdot (\frac{3}{x + 1} + 3)

This expands to

\begin{array}{l} x (x + 1) \cdot \frac{x + 1}{x} + x (x + 1) \cdot 2 \\ = x (x + 1) \cdot \frac{3}{x + 1} + x (x + 1) \cdot 3 \end{array}

x (x + 1) \cdot \frac{x + 1}{x} + x (x + 1) \cdot 2 = x (x + 1) \cdot \frac{3}{x + 1} + x (x + 1) \cdot 3

Now, you can cancel some factors:

\begin{array}{l} x (x + 1) \frac{x + 1}{x} + x (x + 1) 2 \\ = x (x + 1) \frac{3}{x + 1} + x (x + 1) 3 \end{array}

x (x + 1) \frac{x + 1}{x} + x (x + 1) 2 = x (x + 1) \frac{3}{x + 1} + x (x + 1) 3

Now the expression simplifies to

\begin{array}{l} x^{2} + 2 x + 1 + 2 x^{2} + 2 x \\ = 3 x + 3 x^{2} + 3 x \end{array}

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

Isolate all the variables on one side and the constants on the other:

\begin{array}{l} - 1 = x^{2} & + 2 x^{2} - 3 x^{2} \\ + 2 x + 2 x - 3 x - 3 x \end{array}

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

This simplifies to

- 2 x = - 1

Finally, get rid of the number in front of $x$ :

\frac{- 2 x}{- 2} = \frac{- 1}{- 2} = \frac{1}{2}

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

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