Skip to main content
Mathematics LibreTexts

1.8: Complex Fractions

  • Page ID
    40896
  • \( \newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } \)

    \( \newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}} \)

    \( \newcommand{\dsum}{\displaystyle\sum\limits} \)

    \( \newcommand{\dint}{\displaystyle\int\limits} \)

    \( \newcommand{\dlim}{\displaystyle\lim\limits} \)

    \( \newcommand{\id}{\mathrm{id}}\) \( \newcommand{\Span}{\mathrm{span}}\)

    ( \newcommand{\kernel}{\mathrm{null}\,}\) \( \newcommand{\range}{\mathrm{range}\,}\)

    \( \newcommand{\RealPart}{\mathrm{Re}}\) \( \newcommand{\ImaginaryPart}{\mathrm{Im}}\)

    \( \newcommand{\Argument}{\mathrm{Arg}}\) \( \newcommand{\norm}[1]{\| #1 \|}\)

    \( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\)

    \( \newcommand{\Span}{\mathrm{span}}\)

    \( \newcommand{\id}{\mathrm{id}}\)

    \( \newcommand{\Span}{\mathrm{span}}\)

    \( \newcommand{\kernel}{\mathrm{null}\,}\)

    \( \newcommand{\range}{\mathrm{range}\,}\)

    \( \newcommand{\RealPart}{\mathrm{Re}}\)

    \( \newcommand{\ImaginaryPart}{\mathrm{Im}}\)

    \( \newcommand{\Argument}{\mathrm{Arg}}\)

    \( \newcommand{\norm}[1]{\| #1 \|}\)

    \( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\)

    \( \newcommand{\Span}{\mathrm{span}}\) \( \newcommand{\AA}{\unicode[.8,0]{x212B}}\)

    \( \newcommand{\vectorA}[1]{\vec{#1}}      % arrow\)

    \( \newcommand{\vectorAt}[1]{\vec{\text{#1}}}      % arrow\)

    \( \newcommand{\vectorB}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } \)

    \( \newcommand{\vectorC}[1]{\textbf{#1}} \)

    \( \newcommand{\vectorD}[1]{\overrightarrow{#1}} \)

    \( \newcommand{\vectorDt}[1]{\overrightarrow{\text{#1}}} \)

    \( \newcommand{\vectE}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash{\mathbf {#1}}}} \)

    \( \newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } \)

    \(\newcommand{\longvect}{\overrightarrow}\)

    \( \newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}} \)

    \(\newcommand{\ket}[1]{\left| #1 \right>}\)
    \(\newcommand{\bra}[1]{\left< #1 \right|}\)
    \(\newcommand{\braket}[2]{\left< #1 \vphantom{#2} \right| \left. #2 \vphantom{#1} \right>}\)
    \(\newcommand{\braopket}[3]{\left< #1 \vphantom{#2}\vphantom{#3} \right| #2 \vphantom{#1}\vphantom{#3} \left| #3 \vphantom{#1}\vphantom{#2} \right>}\)
    \(\newcommand{\qmvec}[1]{\mathbf{\vec{#1}}}\)
    \(\newcommand{\op}[1]{\hat{\mathbf{#1}}}\)
    \(\newcommand{\expect}[1]{\langle #1 \rangle}\)
    \(\newcommand{\dfn}[1]{\emph{\textbf{#1}}}\)

    \(\newcommand{\avec}{\mathbf a}\) \(\newcommand{\bvec}{\mathbf b}\) \(\newcommand{\cvec}{\mathbf c}\) \(\newcommand{\dvec}{\mathbf d}\) \(\newcommand{\dtil}{\widetilde{\mathbf d}}\) \(\newcommand{\evec}{\mathbf e}\) \(\newcommand{\fvec}{\mathbf f}\) \(\newcommand{\nvec}{\mathbf n}\) \(\newcommand{\pvec}{\mathbf p}\) \(\newcommand{\qvec}{\mathbf q}\) \(\newcommand{\svec}{\mathbf s}\) \(\newcommand{\tvec}{\mathbf t}\) \(\newcommand{\uvec}{\mathbf u}\) \(\newcommand{\vvec}{\mathbf v}\) \(\newcommand{\wvec}{\mathbf w}\) \(\newcommand{\xvec}{\mathbf x}\) \(\newcommand{\yvec}{\mathbf y}\) \(\newcommand{\zvec}{\mathbf z}\) \(\newcommand{\rvec}{\mathbf r}\) \(\newcommand{\mvec}{\mathbf m}\) \(\newcommand{\zerovec}{\mathbf 0}\) \(\newcommand{\onevec}{\mathbf 1}\) \(\newcommand{\real}{\mathbb R}\) \(\newcommand{\twovec}[2]{\left[\begin{array}{r}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\ctwovec}[2]{\left[\begin{array}{c}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\threevec}[3]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\cthreevec}[3]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\fourvec}[4]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\cfourvec}[4]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\fivevec}[5]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\cfivevec}[5]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\mattwo}[4]{\left[\begin{array}{rr}#1 \amp #2 \\ #3 \amp #4 \\ \end{array}\right]}\) \(\newcommand{\laspan}[1]{\text{Span}\{#1\}}\) \(\newcommand{\bcal}{\cal B}\) \(\newcommand{\ccal}{\cal C}\) \(\newcommand{\scal}{\cal S}\) \(\newcommand{\wcal}{\cal W}\) \(\newcommand{\ecal}{\cal E}\) \(\newcommand{\coords}[2]{\left\{#1\right\}_{#2}}\) \(\newcommand{\gray}[1]{\color{gray}{#1}}\) \(\newcommand{\lgray}[1]{\color{lightgray}{#1}}\) \(\newcommand{\rank}{\operatorname{rank}}\) \(\newcommand{\row}{\text{Row}}\) \(\newcommand{\col}{\text{Col}}\) \(\renewcommand{\row}{\text{Row}}\) \(\newcommand{\nul}{\text{Nul}}\) \(\newcommand{\var}{\text{Var}}\) \(\newcommand{\corr}{\text{corr}}\) \(\newcommand{\len}[1]{\left|#1\right|}\) \(\newcommand{\bbar}{\overline{\bvec}}\) \(\newcommand{\bhat}{\widehat{\bvec}}\) \(\newcommand{\bperp}{\bvec^\perp}\) \(\newcommand{\xhat}{\widehat{\xvec}}\) \(\newcommand{\vhat}{\widehat{\vvec}}\) \(\newcommand{\uhat}{\widehat{\uvec}}\) \(\newcommand{\what}{\widehat{\wvec}}\) \(\newcommand{\Sighat}{\widehat{\Sigma}}\) \(\newcommand{\lt}{<}\) \(\newcommand{\gt}{>}\) \(\newcommand{\amp}{&}\) \(\definecolor{fillinmathshade}{gray}{0.9}\)

    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) \(\frac{\left(\frac{1}{2 x-1}-\frac{1}{2 x+1}\right)}{\left(4-\frac{1}{x^{2}}\right)}\)


    1.8: Complex Fractions is shared under a CC BY-NC-SA 4.0 license and was authored, remixed, and/or curated by LibreTexts.

    • Was this article helpful?