ORCCA Complex Fractions

Section 12.4 Complex Fractions

Objectives: PCC Course Content and Outcome Guide

[cross-reference to target(s) "ccog-complex-rational-expressions" missing or not unique]

In this section, we will learn how to simplify complex fractions, which have fractions in the numerator and/or denominator of another fraction.

Figure 12.4.1. Alternative Video Lesson

Subsection 12.4.1 Simplifying Complex Fractions

Consider the rational expression

\begin{equation*} \frac{\frac{6}{x-4}}{\frac{6}{x-4}+3}\text{.} \end{equation*}

It’s difficult to quickly evaluate this expression, or determine the important information such as its domain. This type of rational expression, which contains a “fraction within a fraction,” is referred to as a complex fraction. Our goal is to simplify such a fraction so that it has a single numerator and a single denominator, neither of which contain any fractions themselves.

A complex fraction may have fractions in its numerator and/or denominator. Here is an example to show how we use division to simplify a complex fraction.

\begin{align*} \frac{\ \frac{1}{2}\ }{3}\amp=\frac{1}{2}\div3\\ \amp=\frac{1}{2}\div\frac{3}{1}\\ \amp=\frac{1}{2}\cdot\frac{1}{3}\\ \amp=\frac{1}{6} \end{align*}

What if the expression had something more complicated in the denominator, like \(\frac{\frac{1}{2}}{\frac{1}{3}+\frac{1}{4}}\text{?}\) We would no longer be able to simply multiply by the reciprocal of the denominator, since we don’t immediately know the reciprocal of that denominator. Instead, we could multiply the “main” numerator and denominator by something that eliminates all of the “internal” denominators. (We’ll use the LCD to determine this). For example, with \(\frac{\ \frac{1}{2}\ }{3}\text{,}\) we can multiply by \(\frac{2}{2}\text{:}\)

\begin{align*} \frac{\ \frac{1}{2}\ }{3}\amp=\frac{\ \frac{1}{\cancelhighlight{2}}\ }{3}\multiplyright{\frac{\cancelhighlight{2}}{2}}\\ \amp=\frac{1}{6} \end{align*}

Remark 12.4.2.

In the last example, it’s important to identify which fraction bar is the “main” fraction bar, and which fractions are “internal.” Comparing the two expressions below, both of which are “one over two over three”, we see that they are not equivalent.

\begin{align*} \frac{\ \frac{1}{2}\ }{3}\amp=\frac{\ \frac{1}{\cancelhighlight{2}}\ }{3}\multiplyright{\frac{\cancelhighlight{2}}{2}}\amp\amp\text{versus}\amp\frac{\ 1\ }{\frac{2}{3}}\amp=\frac{\ 1\ }{\frac{2}{\cancelhighlight{3}}}\multiplyright{\frac{3}{\cancelhighlight{3}}}\\ \amp=\frac{1}{6}\amp\amp\amp\amp=\frac{3}{2} \end{align*}

For the first of these, the “main” fraction bar is above the \(3\text{,}\) but for the second of these, the “main” fraction bar is above the \(\frac{2}{3}\text{.}\)

To attack multiple fractions in a complex fraction, we need to multiply the numerator and denominator by the LCD of all the internal fractions, as we will show in the next example.

Example 12.4.3.

Simplify the complex fraction \(\dfrac{\frac{1}{2}}{\frac{1}{3}+\frac{1}{4}}\text{.}\)

Explanation.

The internal denominators are \(2\text{,}\) \(3\text{,}\) and \(4\text{,}\) so the LCD is \(12\text{.}\) We will thus multiply the main numerator and denominator by \(12\) and simplify the result:

\begin{align*} \frac{\frac{1}{2}}{\frac{1}{3}+\frac{1}{4}}\amp=\frac{\frac{1}{2}}{\frac{1}{3}+\frac{1}{4}}\multiplyright{\frac{12}{12}}\\ \amp=\frac{\frac{1}{2}\multiplyright{12}}{\left(\frac{1}{3}+\frac{1}{4}\right)\multiplyright{12}}\\ \amp=\frac{\frac{1}{2}\cdot12}{\frac{1}{3}\cdot12+\frac{1}{4}\cdot12}\\ \amp=\frac{6}{4+3}\\ \amp=\frac{6}{7} \end{align*}

Next we will evaluate a function whose formula is a complex fraction and then simplify the result.

Example 12.4.4.

Find each function value for \(f(x)=\frac{\frac{x+2}{x+3}}{\frac{2}{x+3}-\frac{3}{x-1}}\text{.}\)

\(\displaystyle f(4)\)
\(\displaystyle f(0)\)
\(\displaystyle f(-3)\)
\(\displaystyle f(-11)\)

Explanation.

We will determine each function value by replacing \(x\) with the specified number and then simplify the complex fraction:

\(\displaystyle \begin{aligned}[t] f(\substitute{4})\amp=\frac{\frac{\substitute{4}+2}{\substitute{4}+3}}{\frac{2}{\substitute{4}+3}-\frac{3}{\substitute{4}-1}}\\ \amp=\frac{\frac{6}{7}}{\frac{2}{7}-\frac{3}{3}}\\ \amp=\frac{\frac{6}{7}}{\frac{2}{7}-1}\multiplyright{\frac{7}{7}}\\ \amp=\frac{6}{2-7}\\ \amp=-\frac{6}{5} \end{aligned}\)
\(\displaystyle \begin{aligned}[t] f(\substitute{0})\amp=\frac{\frac{\substitute{0}+2}{\substitute{0}+3}}{\frac{2}{\substitute{0}+3}-\frac{3}{\substitute{0}-1}}\\ \amp=\frac{\frac{2}{3}}{\frac{2}{3}-\frac{3}{-1}}\\ \amp=\frac{\frac{2}{3}}{\frac{2}{3}+3}\multiplyright{\frac{3}{3}}\\ \amp=\frac{2}{2+9}\\ \amp=\frac{2}{11} \end{aligned}\)
When evaluating \(f\) at \(-3\text{,}\) we can quickly see that this results in division by zero:

\begin{align*} f(\substitute{-3})\amp=\frac{\frac{\substitute{-3}+2}{\substitute{-3}+3}}{\frac{2}{\substitute{-3}+3}-\frac{3}{\substitute{-3}-1}}\\ \amp=\frac{\frac{2}{0}}{\frac{2}{0}-\frac{3}{-4}} \end{align*}

Thus \(f(-3)\) is undefined.
\(\begin{aligned}[t] f(\substitute{-11})\amp=\frac{\frac{\substitute{-11}+2}{\substitute{-11}+3}}{\frac{2}{\substitute{-11}+3}-\frac{3}{\substitute{-11}-1}}\\ \amp=\frac{\frac{-8}{-9}}{\frac{2}{-8}-\frac{3}{-12}}\\ \amp=\frac{\frac{8}{9}}{-\frac{1}{4}+\frac{1}{4}}\\ \amp=\frac{\frac{8}{9}}{0} \end{aligned}\)

Therefore \(f(-11)\) is undefined.

We have simplified complex fractions involving numbers and now we will apply the same concept to complex fractions with variables.

Example 12.4.5.

Simplify the complex fraction \(\dfrac{3}{\frac{1}{y}+\frac{5}{y^2}}\text{.}\)

Explanation.

To start, we look at the internal denominators and identify the LCD as \(y^2\text{.}\) We’ll multiply the main numerator and denominator by the LCD, and then simplify. Since we are multiplying by \(\frac{y^2}{y^2}\text{,}\) it is important to note that \(y\) cannot be \(0\text{,}\) since \(\frac{0}{0}\) is undefined.

\begin{align*} \frac{3}{\frac{1}{y}+\frac{5}{y^2}}\amp=\frac{3}{\frac{1}{y}+\frac{5}{y^2}}\multiplyright{\frac{y^2}{y^2}}\\ \amp=\frac{3\multiplyright{y^2}}{\frac{1}{y}\multiplyright{y^2}+\frac{5}{y^2}\multiplyright{y^2}}\\ \amp=\frac{3y^2}{y+5}, \text{ for }y\neq 0 \end{align*}

Example 12.4.6.

Simplify the complex fraction \(\dfrac{\ \frac{5x-6}{2x+1}\ }{\frac{3x+2}{2x+1}}\text{.}\)

Explanation.

The internal denominators are both \(2x+1\text{,}\) so this is the LCD and we will multiply the main numerator and denominator by this expression. Since we are multiplying by \(\frac{2x+1}{2x+1}\text{,}\) what \(x\)-value would cause \(2x+1\) to equal \(0\text{?}\) Solving \(2x+1=0\) leads to \(x=-\frac{1}{2}\text{.}\) So \(x\) cannot be \(-\frac{1}{2}\text{,}\) since \(\frac{0}{0}\) is undefined.

\begin{align*} \frac{\ \frac{5x-6}{2x+1}\ }{\frac{3x+2}{2x+1}}\amp=\frac{\ \frac{5x-6}{\cancelhighlight{2x+1}}\ }{\frac{3x+2}{\secondcancelhighlight{2x+1}}}\multiplyright{\frac{\cancelhighlight{2x+1}}{\secondcancelhighlight{2x+1}}}\\ \amp=\frac{5x-6}{3x+2}, \text{ for }x\neq -\frac{1}{2} \end{align*}

Example 12.4.7.

Completely simplify the function defined by \(f(x)=\frac{\frac{x+2}{x+3}}{\frac{2}{x+3}-\frac{3}{x-1}}\text{.}\) Then determine the domain of this function.

Explanation.

The LCD of the internal denominators is \((x+3)(x-1)\text{.}\) We will thus multiply the main numerator and denominator by the expression \((x+3)(x-1)\) and then simplify the resulting expression.

\begin{align*} f(x)\amp=\frac{\frac{x+2}{x+3}}{\frac{2}{x+3}-\frac{3}{x-1}}\\ \amp=\frac{\frac{x+2}{x+3}}{\frac{2}{x+3}-\frac{3}{x-1}}\multiplyright{\frac{(x+3)(x-1)}{(x+3)(x-1)}}\\ \amp=\frac{\frac{x+2}{x+3}\multiplyright{(x+3)(x-1)}}{\left(\frac{2}{x+3}-\frac{3}{x-1}\right)\multiplyright{(x+3)(x-1)}}\\ \amp=\frac{\frac{x+2}{\cancelhighlight{x+3}}\cancelhighlight{(x+3)}(x-1)}{\frac{2}{\secondcancelhighlight{x+3}}\secondcancelhighlight{(x+3)}(x-1)-\frac{3}{\thirdcancelhighlight{x-1}}(x+3)\thirdcancelhighlight{(x-1)}}\\ \amp=\frac{(x+2)(x-1)}{2(x-1)-3(x+3)}, \text{ for }x\neq -3, x\neq 1\\ \amp=\frac{(x+2)(x-1)}{2x-2-3x-9}, \text{ for }x\neq -3, x\neq 1\\ \amp=\frac{(x+2)(x-1)}{-x-11}, \text{ for }x\neq -3, x\neq 1\\ \amp=\frac{(x+2)(x-1)}{-(x+11)}, \text{ for }x\neq -3, x\neq 1 \end{align*}

In the original (unsimplified) function, we could see that \(x\neq -3\) and \(x\neq 1\text{.}\) In the simplified function, we need \(x+11\neq 0\text{,}\) so we can also see that \(x\neq -11\text{.}\) Therefore the domain of the function \(f\) is \(\{x\mid x\neq -11, -3, 1\}\text{.}\)

Example 12.4.8.

Simplify the complex fraction \(\dfrac{2\left(\frac{-4x+3}{x-2}\right)+3}{\frac{-4x+3}{x-2}+4}\text{.}\)

Explanation.

The only internal denominator is \(x-2\text{,}\) so we will begin by multiplying the main numerator and denominator by this. Then we’ll simplify the resulting expression.

\begin{align*} \frac{2\left(\frac{-4x+3}{x-2}\right)+3}{\frac{-4x+3}{x-2}+4}\amp=\frac{2\left(\frac{-4x+3}{x-2}\right)+3}{\frac{-4x+3}{x-2}+4}\multiplyright{\frac{x-2}{x-2}}\\ \amp=\frac{2\left(\frac{-4x+3}{\cancelhighlight{x-2}}\right)\cancelhighlight{(x-2)}+3(x-2)}{\left(\frac{-4x+3}{\secondcancelhighlight{x-2}}\right)\secondcancelhighlight{(x-2)}+4(x-2)}\\ \amp=\frac{2(-4x+3)+3(x-2)}{(-4x+3)+4(x-2)}, \text{ for }x\neq 2\\ \amp=\frac{-8x+6+3x-6}{-4x+3+4x-8}, \text{ for }x\neq 2\\ \amp=\frac{-5x}{-5}, \text{ for }x\neq 2\\ \amp=x, \text{ for } x\neq 2 \end{align*}

Example 12.4.9.

Simplify the complex fraction \(\dfrac{\frac{5}{x}+\frac{4}{y}}{\frac{3}{x}-\frac{2}{y}}\text{.}\) Recall that with a multivariable expression, this textbook ignores domain restrictions.

Explanation.

We multiply the numerator and denominator by the common denominator of \(x\) and \(y\text{,}\) which is \(xy\text{:}\)

\begin{align*} \frac{\frac{5}{x}+\frac{4}{y}}{\frac{3}{x}-\frac{2}{y}}\amp=\frac{\frac{5}{x}+\frac{4}{y}}{\frac{3}{x}-\frac{2}{y}}\multiplyright{\frac{xy}{xy}}\\ \amp=\frac{\left(\frac{5}{x}+\frac{4}{y}\right)xy}{\left(\frac{3}{x}-\frac{2}{y}\right)xy}\\ \amp=\frac{\frac{5}{\cancelhighlight{x}}\cancelhighlight{x}y+\frac{4}{\secondcancelhighlight{y}}x\secondcancelhighlight{y}}{\frac{3}{\cancelhighlight{x}}\cancelhighlight{x}y-\frac{2}{\secondcancelhighlight{y}}x\secondcancelhighlight{y}}\\ \amp=\frac{5y+4x}{3y-2x} \end{align*}

Example 12.4.10.

Simplify the complex fraction \(\dfrac{\frac{t}{t+3}+\frac{2}{t-3}}{1-\frac{t}{t^2-9}}\text{.}\)

Explanation.

First, we check all quadratic polynomials to see if they can be factored and factor them:

\begin{equation*} \frac{\frac{t}{t+3}+\frac{2}{t-3}}{1-\frac{t}{t^2-9}}=\frac{\frac{t}{t+3}+\frac{2}{t-3}}{1-\frac{t}{(t-3)(t+3)}} \end{equation*}

Next, we identify the common denominator of the three fractions, which is \((t+3)(t-3)\text{.}\) We then multiply the main numerator and denominator by that expression:

\begin{align*} \frac{\frac{t}{t+3}+\frac{2}{t-3}}{1-\frac{t}{t^2-9}}\amp=\frac{\frac{t}{t+3}+\frac{2}{t-3}}{1-\frac{t}{(t-3)(t+3)}}\multiplyright{\frac{(t+3)(t-3)}{(t+3)(t-3)}}\\ \amp=\frac{\frac{t}{\cancelhighlight{t+3}}\cancelhighlight{(t+3)}(t-3)+\frac{2}{\secondcancelhighlight{t-3}}(t+3)\secondcancelhighlight{(t-3)}}{1(t+3)(t-3)-\frac{t}{\thirdcancelhighlight{(t-3)(t+3)}}\thirdcancelhighlight{(t+3)(t-3)}}\\ \amp=\frac{t(t-3)+2(t+3)}{(t+3)(t-3)-t} \text{ for }t\neq -3, t\neq 3\\ \amp=\frac{t^2-3t+2t+6}{t^2-9-t} \text{ for }t\neq -3, t\neq 3\\ \amp=\frac{t^2-t+6}{t^2-t-9} \text{ for }t\neq -3, t\neq 3 \end{align*}

Note that since both the numerator and denominator are prime trinomials, this expression can neither factor nor simplify any further.

Reading Questions 12.4.2 Reading Questions

1.

What does it mean for a fraction to be a “complex” fraction?

2.

When simplifying a complex fraction, why is it necessary to keep track of domain restrictions?

Exercises 12.4.3 Exercises

Review and Warmup.

1.

Calculate the following. Use an improper fraction in your answer.

\(\displaystyle{\frac{ \frac{25}{4} }{ \frac{5}{9} } =}\)
\(\displaystyle{\frac{ \frac{x}{r} }{ \frac{t}{y} } =}\)

2.

Calculate the following. Use an improper fraction in your answer.

\(\displaystyle{\frac{ \frac{5}{3} }{ \frac{5}{4} } =}\)
\(\displaystyle{\frac{ \frac{y}{x} }{ \frac{t}{r} } =}\)

3.

Calculate the following. Use an improper fraction in your answer.

\(\displaystyle{\frac{ 5 }{ \frac{4}{9} } =}\)
\(\displaystyle{\frac{ \frac{5}{4} }{9} =}\)

4.

Calculate the following. Use an improper fraction in your answer.

\(\displaystyle{\frac{ 3 }{ \frac{4}{5} } =}\)
\(\displaystyle{\frac{ \frac{3}{4} }{5} =}\)

5.

Calculate the following. Use an improper fraction in your answer.

\(\displaystyle{\frac{ \frac{2}{5} + \frac{2}{3} }{ \frac{3}{4} } =}\)

6.

Calculate the following. Use an improper fraction in your answer.

\(\displaystyle{\frac{ \frac{3}{5} - \frac{1}{6} }{ \frac{4}{3} } =}\)

7.

Calculate the following. Use an improper fraction in your answer.

\(\displaystyle{\frac{6}{ \frac{4}{3} + \frac{2}{3} } =}\)

8.

Calculate the following. Use an improper fraction in your answer.

\(\displaystyle{\frac{1}{ \frac{2}{5} - \frac{2}{3} } =}\)