Skip to main content

Section 20.5 Going Deeper: Fractions Inside of Fractions

The technique of multiplying the top and bottom of a fraction by the same value to eliminate the denominators seems odd when working with fractions involving numbers, but its value goes up significantly when working with rational expressions. One of the more algebraically difficult situations to work with is when you have fractions inside of other fractions. It's not that anything about the process is different, but the sheer number of symbols can cause students to feel overwhelmed and make mistakes.

Here is an example. The goal is to simplify the following fraction:

\begin{equation*} \frac{ \frac{1}{x + 1} - \frac{3}{x - 2}}{ \frac{2} {x - 2} + \frac{1}{x + 3}} \end{equation*}

We will first do the calculation by adding the fractions in the numerator and denominator, and then dividing the resulting fractions by multiplying by the reciprocal.

\begin{equation*} \begin{aligned} \frac{\frac{1}{x + 1} - \frac{3}{x - 2}}{ \frac{2} {x - 2} + \frac{1}{x + 3}} \amp = \frac{ \frac{1 \cdot (x - 2)}{(x + 1) \cdot (x - 2)} - \frac{3 \cdot (x + 1)}{(x - 2) \cdot (x + 1)}}{ \frac{2 \cdot (x + 3)} {(x - 2) \cdot (x + 3)} + \frac{1 \cdot (x - 2)}{(x + 3) \cdot (x - 2)}} \amp \eqnspacer \amp \text{Common Denominator} \\ \amp = \frac{ \frac{(x - 2) - (3x + 3)}{(x + 1) (x - 2)} }{ \frac{(2x + 6) + (x - 2)} {(x - 2) (x + 3)}} \\ \amp = \frac{ \frac{-2x - 5}{(x + 1) (x - 2)} }{ \frac{3x + 4} {(x - 2) (x + 3)}} \amp \amp \text{Combine like terms} \\ \amp = \frac{-2x - 5}{(x + 1) (x - 2)} \cdot \frac{(x - 2) (x + 3)}{3x + 4} \amp \amp \text{Multiply by the reciprocal} \\ \amp = - \frac{(2x + 5)(x - 2)(x + 3)}{(x + 1)(x - 2)(3x + 4)} \amp \amp \text{Multiply and factor out the negative} \\ \amp = - \frac{(2x + 5)(x + 3) \cdot \cancel{(x - 2)}}{(x + 1)(3x + 4) \cdot \cancel{(x - 2)}} \amp \amp \text{Reduce} \\ \amp = - \frac{(2x + 5)(x + 3)}{(x + 1)(3x + 4)} \amp \amp \text{Reduce} \\ \end{aligned} \end{equation*}

We will now do the same calculation, but we will multiply the numerator and denominator by the same value to eliminate the fractions inside of the fraction.

\begin{equation*} \begin{aligned} \frac{\frac{1}{x + 1} - \frac{3}{x - 2}}{ \frac{2} {x - 2} + \frac{1}{x + 3}} \amp = \frac{\frac{1}{x + 1} - \frac{3}{x - 2}}{ \frac{2} {x - 2} + \frac{1}{x + 3}} \cdot \frac{(x+1)(x-2)(x+3)}{(x+1)(x-2)(x+3)} \amp \eqnspacer \amp \text{Multiply top and bottom} \\ \amp = \frac{\frac{(x+1)(x-2)(x+3)}{x + 1} - \frac{3(x+1)(x-2)(x+3)}{x - 2}}{ \frac{2(x+1)(x-2)(x+3)} {x - 2} + \frac{(x+1)(x-2)(x+3)}{x + 3}} \amp \amp \text{Distribute} \\ \amp = \frac{\frac{\cancel{(x+1)}(x-2)(x+3)}{\cancel{x + 1}} - \frac{3(x+1)\cancel{(x-2)}(x+3)}{\cancel{x - 2}}}{ \frac{2(x+1)\cancel{(x-2)}(x+3)} {\cancel{x - 2}} + \frac{(x+1)(x-2)\cancel{(x+3)}}{\cancel{x + 3}}} \amp \amp \text{Reduce} \\ \amp = \frac{(x-2)(x+3) - 3(x+1)(x+3)}{2(x+1)(x+3) + (x+1)(x-2)} \amp \amp \text{Rewrite and simplify} \\ \amp = \frac{(x^2 + x - 6) - (3x^2 + 12x + 9)}{(2x^2 + 8x + 6) + (x^2 - x - 2)} \\ \amp = \frac{-2x^2 - 11 x - 15}{3x^2 + 7x + 4} \\ \amp = - \frac{2x^2 + 11 x + 15}{3x^2 + 7x + 4} \amp \amp \text{Factor out the negative} \\ \amp = - \frac{(2x + 5)(x + 3)}{(x+1)(3x + 4)} \end{aligned} \end{equation*}

Both of these calculations involve a lot of steps, and neither one is necessarily better than the other. The first one required the proper execution of more fraction calculations, but the second one required a factorization step to ensure that the expression cannot be reduced. Both of those steps can be difficult in their own ways.

It should be noted that expressions that are this messy do not appear particularly frequently, but if you continue along in mathematics, you will find that calculations involving fractions inside of fractions do appear. There are two reasons for including this calculation here. The first is to expose you to some of the more intricate algebraic manipulations that arise in higher levels of mathematics. The second is to point out the fact that even though the algebra may look complicated, there's nothing here that you haven't seen before or is beyond your ability. Your capacity for handling complicated expressions like these will grow as you continue to practice and gain experience.