Subsection13.4.1Simplifying Complex Fractions
A complex fraction has fractions in its numerator and/or denominator, like
\begin{equation*}
\frac{\frac{1}{2}}{3}
\end{equation*}
We need to remove fractions from the numerator and denominator. Here is the normal method to simplify this problem:
\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*}
We will learn an easier method. In a fraction, we multiply the numerator and denominator by the same number, and the fraction's value doesn't change. For example:
\begin{align*}
\frac{1}{2}\amp=\frac{1\multiplyright{2}}{2\multiplyright{2}}=\frac{2}{4}\\
\frac{1}{2}\amp=\frac{1\multiplyright{3}}{2\multiplyright{3}}=\frac{3}{6}
\end{align*}
Note that \(\frac{1}{2}\cdot2=1\text{.}\) For \(\frac{\frac{1}{2}}{3}\text{,}\) to remove the fraction \(\frac{1}{2}\text{,}\) we multiply the numerator and denominator by \(2\text{,}\) and we have:
\begin{align*}
\frac{\frac{1}{2}}{3}\amp=\frac{\frac{1}{2}\multiplyright{2}}{3\multiplyright{2}}\\
\amp=\frac{1}{6}
\end{align*}
Compare the problem above with the following one:
\begin{align*}
\frac{1}{\frac{2}{3}}\amp=\frac{{1}\multiplyright{3}}{\frac{2}{3}\multiplyright{3}}\\
\amp=\frac{3}{2}
\end{align*}
So \(\frac{\frac{1}{2}}{3}\) and \(\frac{1}{\frac{2}{3}}\) have different values.
To remove multiple fractions in a complex fraction, we need to multiply the numerator and denominator by those fractions' common denominator, like in the next few examples.
Example13.4.2
Simplify \(\frac{\frac{1}{2}}{\frac{1}{3}}\)
SolutionWe multiply the numerator and denominator by the common denominator of \(2\) and \(3\text{,}\) which is \(6\text{:}\)
\begin{align*}
\frac{\frac{1}{2}}{\frac{1}{3}}\amp=\frac{\frac{1}{2}\multiplyright{6}}{\frac{1}{3}\multiplyright{6}}\\
\amp=\frac{3}{2}
\end{align*}
Since the fraction line serves the same function as the division sign, we have the following shortcut when we multiply fractions:
\begin{align*}
\frac{1}{2}\cdot{6}\amp=6\div2\cdot1=3\\
\frac{1}{3}\cdot{6}\amp=6\div3\cdot1=2
\end{align*}
Example13.4.3
Simplify \(\frac{\frac{1}{2}+\frac{1}{3}}{5}\)
SolutionWe need to remove \(\frac{1}{2}\) and \(\frac{1}{3}\text{,}\) so we multiply the numerator and denominator by the common denominator of \(2\) and \(3\text{,}\) which is \(6\text{.}\) Note that we don't consider \(5\) when we calculate the common denominator, because \(5\) is not a fraction. The solution is:
\begin{align*}
\frac{\frac{1}{2}+\frac{1}{3}}{5}\amp=\frac{\multiplyleft{6}(\frac{1}{2}+\frac{1}{3})}{\multiplyleft{6}5}\\
\amp=\frac{6\cdot\frac{1}{2}+6\cdot\frac{1}{3}}{30}\\
\amp=\frac{3+2}{30}\\
\amp=\frac{5}{30}\\
\amp=\frac{1}{6}
\end{align*}
The same concept can be applied to fractions with variables, like in the next few examples.
Example13.4.4
Simplify \(\frac{\frac{1}{x}+\frac{1}{y}}{\frac{1}{x}-\frac{1}{y}}\)
SolutionWe multiply the numerator and denominator by the common denominator of \(x\) and \(y\text{,}\) which is \(xy\text{:}\)
\begin{align*}
\frac{\frac{1}{x}+\frac{1}{y}}{\frac{1}{x}-\frac{1}{y}}\amp=\frac{\multiplyleft{xy}(\frac{1}{x}+\frac{1}{y})}{\multiplyleft{xy}(\frac{1}{x}-\frac{1}{y})}\\
\amp=\frac{xy\cdot\frac{1}{x}+xy\cdot\frac{1}{y}}{xy\cdot\frac{1}{x}-xy\cdot\frac{1}{y}}\\
\amp=\frac{y+x}{y-x}
\end{align*}
Example13.4.5
Simplify \(\frac{\frac{1}{x+3}+\frac{1}{x-3}}{\frac{1}{x^2-9}-1}\)
SolutionFirst, we factor all denominators if possible:
\begin{equation*}
\frac{\frac{1}{x+3}+\frac{1}{x-3}}{\frac{1}{x^2-9}-1}=\frac{\frac{1}{x+3}+\frac{1}{x-3}}{\frac{1}{(x+3)(x-3)}-1}
\end{equation*}
Next, identify the common denominator of those three fractions, which is \((x+3)(x-3)\text{.}\) We multiply the numerator and denominator by the common denominator:
\begin{align*}
\frac{\frac{1}{x+3}+\frac{1}{x-3}}{\frac{1}{x^2-9}-1}\amp=\frac{\frac{1}{x+3}+\frac{1}{x-3}}{\frac{1}{(x+3)(x-3)}-1}\\
\amp=\frac{\multiplyleft{(x+3)(x-3)}[\frac{1}{x+3}+\frac{1}{x-3}]}{\multiplyleft{(x+3)(x-3)}[\frac{1}{(x+3)(x-3)}-1]}\\
\amp=\frac{(x+3)(x-3)\cdot\frac{1}{x+3}+(x+3)(x-3)\cdot\frac{1}{x-3}}{(x+3)(x-3)\cdot\frac{1}{(x+3)(x-3)}-(x+3)(x-3)\cdot1}\\
\amp=\frac{(x-3)+(x+3)}{1-(x+3)(x-3)}\\
\amp=\frac{x-3+x+3}{1-(x^2-9)}\\
\amp=\frac{2x}{1-x^2+9}\\
\amp=\frac{2x}{10-x^2}
\end{align*}
