Skip to main content
Mathematics LibreTexts

Complex Fractions and Equations with Rational Expressions

  • Page ID
    221404
  • \( \newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } \)

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

    Complex Fractions and Equations with Rational Expressions

    Complex Fractions

    First we begin with a complex fraction that contains no variables.

    Example

           1         5                          1                  5
                 -                                     12   -            12    
           2         6                          2                  6       
                                        =                                               
    Multiply Numerator and 
           1          2                           1                 2                 Denominator by 12
                 +                                      12  +           12 
           4          3                           4                 3

     

                6 - 10                    4  
        =                      =     -          
                3 + 8                    11 

     

    Notice we first multiplied by the total Least Common Denominator, then we simplified.

     


    Complex Fractions Involving Expressions

    When we have a complex fraction with rational expressions as the numerator and denominator, we follow similar steps, except, of course factoring plays a key role.  

     

    • Step 1  Factor everything.
       

    • Step 2  Determine the total least common denominator, using the maximum power of each factor.
       

    • Step 3  Multiply all terms by the LCD.
       

    • Step 4  Combine like terms.
       

    • Step 5  Factor and Cancel.

     

    Note:  Usually you will not have to do all of the steps.

     

    Example:  
     

                   7                                                7
          1 -                                   1(x + 1)  -             (x + 1)    
                x + 1                                           x + 1       
                                        =                                                   
    Multiply Numerator and 
             4                                    4                                         Denominator by (x+1)
                    +   1                               (x + 1)  + 1(x + 1) 
          x + 1                              x + 1            

     

                x + 1 - 7              x - 6 
        =                        =                 
               4 + x + 1              x + 5 

     


    Cross Multiplication

    Recall that if 

             a          c
                   =        
             b          d

    then 

            ad  =  bc  

    The same hold true for functions:

             f           g
                   =        
             h           k

    then 

            fg   =   hk 

     

    Example

    Solve.

             3x - 1           x + 2
                          =                
             5x - 2          3x + 4

     

    Solution

    We cross multiply

            (3x - 1)(3x + 4)  =  (5x - 2)(x + 2)

            9x2 + 12x - 3x - 4  =  5x2 + 10x - 2x - 4

            9x2 + 9x - 4  =  5x2 + 8x - 4

            4x2 + x  =  0

            x (4x + 1)  =  0

                                           1
            x = 0     
    or     x = -      
                                           4

     

    Caution:  Always check and see that the solution works by plugging back into the original equation!

     


    Equations with Rational Expressions

    To solve equations that involve rational expressions, we following the following steps:

     

    • Step 1  Factor if possible.
       

    • Step 2  Multiply the left hand and right hand sides by the LCD.
       

    • Step 3  Combine like terms.
       

    • Step 4  Bring everything to the left side of the equation.
       

    • Step 5  Solve by the zero product method or by basic algebra.
       

    • Step 6  Plug back in to the original equation to check for extraneous solutions.

     

    Example

    Solve

               3             4                 48
                       -              =               
            x - 6        x + 6          x2 - 36

    Solution

    First factor.

               3               4                   48
                        -              =                       
             x - 6         x + 6        (x - 6)(x + 6)

    Then multiply by the LCD (x - 6)(x+ 6).

               3                                4                                        48
                     (x - 6)(x + 6)  -             (x - 6)(x + 6)  =                        (x - 6)(x + 6)
            x - 6                           x + 6                             (x - 6)(x + 6)

     

            3(x + 6) - 4(x - 6)  =  48

            3x + 18 - 4x + 24  =  48

            -x + 42  =  48

            -x = 6

            x = -6

    Notice that -6 cannot be put back into the original equation, since there would be a zero in the denominator. We can conclude that this equation has no solution.

     


    Back to the Absolute Values, Factoring and Arithmetic of Rational Expressions Page

     

     

    Complex Fractions and Equations with Rational Expressions is shared under a not declared license and was authored, remixed, and/or curated by LibreTexts.

    • Was this article helpful?