1.8: Complex Fractions
- Page ID
- 40896
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\(\newcommand{\avec}{\mathbf a}\) \(\newcommand{\bvec}{\mathbf b}\) \(\newcommand{\cvec}{\mathbf c}\) \(\newcommand{\dvec}{\mathbf d}\) \(\newcommand{\dtil}{\widetilde{\mathbf d}}\) \(\newcommand{\evec}{\mathbf e}\) \(\newcommand{\fvec}{\mathbf f}\) \(\newcommand{\nvec}{\mathbf n}\) \(\newcommand{\pvec}{\mathbf p}\) \(\newcommand{\qvec}{\mathbf q}\) \(\newcommand{\svec}{\mathbf s}\) \(\newcommand{\tvec}{\mathbf t}\) \(\newcommand{\uvec}{\mathbf u}\) \(\newcommand{\vvec}{\mathbf v}\) \(\newcommand{\wvec}{\mathbf w}\) \(\newcommand{\xvec}{\mathbf x}\) \(\newcommand{\yvec}{\mathbf y}\) \(\newcommand{\zvec}{\mathbf z}\) \(\newcommand{\rvec}{\mathbf r}\) \(\newcommand{\mvec}{\mathbf m}\) \(\newcommand{\zerovec}{\mathbf 0}\) \(\newcommand{\onevec}{\mathbf 1}\) \(\newcommand{\real}{\mathbb R}\) \(\newcommand{\twovec}[2]{\left[\begin{array}{r}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\ctwovec}[2]{\left[\begin{array}{c}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\threevec}[3]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\cthreevec}[3]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\fourvec}[4]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\cfourvec}[4]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\fivevec}[5]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\cfivevec}[5]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\mattwo}[4]{\left[\begin{array}{rr}#1 \amp #2 \\ #3 \amp #4 \\ \end{array}\right]}\) \(\newcommand{\laspan}[1]{\text{Span}\{#1\}}\) \(\newcommand{\bcal}{\cal B}\) \(\newcommand{\ccal}{\cal C}\) \(\newcommand{\scal}{\cal S}\) \(\newcommand{\wcal}{\cal W}\) \(\newcommand{\ecal}{\cal E}\) \(\newcommand{\coords}[2]{\left\{#1\right\}_{#2}}\) \(\newcommand{\gray}[1]{\color{gray}{#1}}\) \(\newcommand{\lgray}[1]{\color{lightgray}{#1}}\) \(\newcommand{\rank}{\operatorname{rank}}\) \(\newcommand{\row}{\text{Row}}\) \(\newcommand{\col}{\text{Col}}\) \(\renewcommand{\row}{\text{Row}}\) \(\newcommand{\nul}{\text{Nul}}\) \(\newcommand{\var}{\text{Var}}\) \(\newcommand{\corr}{\text{corr}}\) \(\newcommand{\len}[1]{\left|#1\right|}\) \(\newcommand{\bbar}{\overline{\bvec}}\) \(\newcommand{\bhat}{\widehat{\bvec}}\) \(\newcommand{\bperp}{\bvec^\perp}\) \(\newcommand{\xhat}{\widehat{\xvec}}\) \(\newcommand{\vhat}{\widehat{\vvec}}\) \(\newcommand{\uhat}{\widehat{\uvec}}\) \(\newcommand{\what}{\widehat{\wvec}}\) \(\newcommand{\Sighat}{\widehat{\Sigma}}\) \(\newcommand{\lt}{<}\) \(\newcommand{\gt}{>}\) \(\newcommand{\amp}{&}\) \(\definecolor{fillinmathshade}{gray}{0.9}\)1.8 Complex Fractions
Complex fractions involve simplifying a rational expression which has a complicated numerator and/or denominator
Example
Simplify.
\[
\frac{3+\frac{x}{x+2}}{1-\frac{x+3}{x-1}}
\]
There are a variety of ways to approach this problem. One of the most straightforward ways to simplify the expression above is to create common denominators for the numerator and the denominator so that each one is a single fractional expression:
\[
\begin{aligned}
\frac{3+\frac{x}{x+2}}{1-\frac{x+3}{x-1}} &=\frac{\frac{3}{1} * \frac{x+2}{x+2}+\frac{x}{x+2}}{\frac{1}{1} * \frac{x-1}{x-1}-\frac{x+3}{x-1}} \\
&=\frac{\left(\frac{3 x+6+x}{x+2}\right)}{\left(\frac{x-1-(x+3)}{x-1}\right)}
\end{aligned}
\]
\(=\frac{\left(\frac{4 x+6}{x+2}\right)}{\left(\frac{-4}{x-1}\right)} \quad\) (Now this is a division problem)
\[
\begin{array}{l}
=\frac{4 x+6}{x+2} * \frac{x-1}{-4}=\frac{2(2 x+3)}{x+2} * \frac{x-1}{-4} \\
=\frac{\cancel{2}(2 x+3)}{x+2} * \frac{x-1}{\cancel{-4}(-2)}=\frac{(2 x+3)(x-1)}{-2(x+2)}
\end{array}
\]
Simplifying complex fractions uses all of the previous concepts about rational expressions which we've covered in this chapter.
Example
Simplify.
\[
\frac{x-\frac{x}{x+3}}{1+\frac{2}{x}}
\]
\(\frac{x-\frac{x}{x+3}}{1+\frac{2}{x}}=\frac{\frac{x}{1} * \frac{x+3}{x+3}-\frac{x}{x+3}}{\frac{1}{1} * \frac{x}{x}+\frac{2}{x}} \quad\) creating common denominators
\[
\begin{array}{l}
=\frac{\left(\frac{x(x+3)-x}{x+3}\right)}{\left(\frac{x+2}{x}\right)} \\
=\frac{\left(\frac{x^{2}+3 x-x}{x+3}\right)}{\left(\frac{x+2}{x}\right)}=\frac{\left(\frac{x^{2}+2 x}{x+3}\right)}{\left(\frac{x+2}{x}\right)} \text { dividing fractions } \\
=\frac{x^{2}+2 x}{x+3} * \frac{x}{x+2}=\frac{x(x+2)}{x+3} * \frac{x}{x+2}
\end{array}
\]
\(=\frac{x\cancel{(x+2)}}{x+3} * \frac{x}{\cancel{x+2}} \quad\) factor and cancel to reduce to lowest terms
\[
=\frac{x^{2}}{x+3}
\]
Exercises 1.8
Simplify each complex fraction. Express your answer in lowest terms.
1) \(\frac{1}{\left(x+\frac{y}{2}\right)}\)
2) \(\frac{\left(\frac{1}{x}+\frac{1}{y}\right)}{\left(\frac{y}{x}-\frac{x}{y}\right)}\)
3) \(\frac{\left(1+\frac{m}{n}\right)}{\left(1-\frac{n^{2}}{m^{2}}\right)}\)
4) \(\frac{\left(\frac{1}{x}-\frac{1}{y}\right)}{\left(\frac{1}{x^{2}}-\frac{1}{y^{2}}\right)}\)
5) \(\frac{\left(\frac{x}{y}-\frac{x-y}{x+y}\right)}{\left(\frac{y}{x}+\frac{x-y}{x+y}\right)}\)
6) \(\frac{\left(\frac{7}{a+1}-\frac{3}{a}\right)}{\left(\frac{3}{a}+\frac{1}{a-1}\right)}\)
7) \(\frac{\left(x-\frac{1}{2 x+1}\right)}{\left(1-\frac{2}{2 x+1}\right)}\)
8) \(\frac{\left(\frac{1}{2 x-2}-\frac{1}{x}\right)}{\left(\frac{2}{x}-\frac{1}{x-1}\right)}\)
9) \(\frac{\left(x+\frac{4}{x+4}\right)}{\left(x-\frac{4 x+4}{x+4}\right)}\)
10) \(\frac{\left(x-\frac{x+6}{x+2}\right)}{\left(x-\frac{4 x+15}{x+2}\right)}\)
11) \(\frac{\left(\frac{1}{x+2}-\frac{1}{x-3}\right)}{\left(1+\frac{1}{x^{2}-x-6}\right)}\)
\(\frac{\left(1-\frac{1}{x+1}\right)}{\left(1+\frac{1}{x-1}\right)}\)
13) \(\frac{\left(\frac{1}{a-b}-\frac{3}{a+b}\right)}{\left(\frac{2}{b-a}+\frac{4}{b+a}\right)}\)
14) \(\frac{\left(\frac{3}{y^{2}-4}\right)}{\left(\frac{1}{y+2}-\frac{1}{y-2}\right)}\)
\(\frac{\left(n+2-\frac{5}{n-2}\right)}{\left(1-\frac{1}{(n-2)^{2}}\right)}\)
16) \(\frac{\left(4+\frac{1}{x+1}\right)}{\left(16-\frac{1}{(x+1)^{2}}\right)}\)
17) \(\frac{\left(2+\frac{x-2}{1-x^{2}}\right)}{\left(2-\frac{3}{x+1}\right)}\)
18) \(\frac{\left(\frac{1}{2 x-1}-\frac{1}{2 x+1}\right)}{\left(4-\frac{1}{x^{2}}\right)}\)