Skip to main content
Mathematics LibreTexts

1.7: Adding and Subtracting Rational Expressions

  • Page ID
    40895
  • \( \newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } \) \( \newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}} \)\(\newcommand{\id}{\mathrm{id}}\) \( \newcommand{\Span}{\mathrm{span}}\) \( \newcommand{\kernel}{\mathrm{null}\,}\) \( \newcommand{\range}{\mathrm{range}\,}\) \( \newcommand{\RealPart}{\mathrm{Re}}\) \( \newcommand{\ImaginaryPart}{\mathrm{Im}}\) \( \newcommand{\Argument}{\mathrm{Arg}}\) \( \newcommand{\norm}[1]{\| #1 \|}\) \( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\) \( \newcommand{\Span}{\mathrm{span}}\) \(\newcommand{\id}{\mathrm{id}}\) \( \newcommand{\Span}{\mathrm{span}}\) \( \newcommand{\kernel}{\mathrm{null}\,}\) \( \newcommand{\range}{\mathrm{range}\,}\) \( \newcommand{\RealPart}{\mathrm{Re}}\) \( \newcommand{\ImaginaryPart}{\mathrm{Im}}\) \( \newcommand{\Argument}{\mathrm{Arg}}\) \( \newcommand{\norm}[1]{\| #1 \|}\) \( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\) \( \newcommand{\Span}{\mathrm{span}}\)\(\newcommand{\AA}{\unicode[.8,0]{x212B}}\)

    Just as we do with numerical fractions, we will need to have common denominators in order to add or subtract algebraic fractions. When we add \(\frac{1}{2}+\frac{1}{3},\) we make a common denominator of 6 so that we can add them together.
    \[
    \begin{aligned}
    \frac{1}{2}+\frac{1}{3} &=\frac{1}{2} * \frac{3}{3}+\frac{1}{3} * \frac{2}{2} \\
    &=\frac{3}{6}+\frac{2}{6} \\
    &=\frac{5}{6}
    \end{aligned}
    \]
    Because the denominators, 2 and \(3,\) are prime and don't share any common factors, the common denominator is simply \(3 * 2=6 .\) We can see a similar result in adding algebraic fractions.

    Example
    Add the fractions. Express your answer in lowest terms.
    \[
    \begin{aligned}
    \frac{2}{x}+\frac{x}{x-3} & \\
    \frac{2}{x}+\frac{x}{x-3} &=\frac{2}{x} * \frac{x-3}{x-3}+\frac{x}{x-3} * \frac{x}{x} \\
    &=\frac{2(x-3)}{x(x-3)}+\frac{x * x}{x(x-3)} \\
    &=\frac{2 x-6+x^{2}}{x(x-3)}=\frac{x^{2}+2 x-6}{x(x-3)}
    \end{aligned}
    \]

    It's important to be aware that in subtraction, the negative sign representing subtraction must be distributed to all terms in the second numerator.
    Example
    Subtract the given expressions. Express your answer in lowest terms.
    \[
    \begin{aligned}
    \frac{6}{x+1}-\frac{x+5}{x-2} \\
    \frac{6}{x+1}-\frac{x+5}{x-2} &=\frac{6}{x+1} * \frac{x-2}{x-2}-\frac{x+5}{x-2} * \frac{x+1}{x+1} \\
    &=\frac{6(x-2)-(x+5)(x+1)}{(x+1)(x-2)} \\
    &=\frac{6 x-12-\left(x^{2}+6 x+5\right)}{(x+1)(x-2)} \\
    &=\frac{6 x-12-x^{2}-6 x-5}{(x+1)(x-2)} \\
    &=\frac{-x^{2}-17}{(x+1)(x-2)}
    \end{aligned}
    \]
    In other situations, the denominators may share a common factor. In this case, we can turn one of the denominators into the other one:

    Example
    Add the given fractions.
    \(\frac{7}{x^{2}+8 x+15}+\frac{2}{x+3}\)

    \[
    \frac{7}{x^{2}+8 x+15}+\frac{2}{x+3}=\frac{7}{(x+3)(x+5)}+\frac{2}{x+3}
    \]
    We can turn \((x+3)\) into \(x^{2}+8 x+15\) by multiplying by \((x+5)\)
    \[
    \begin{aligned}
    \frac{7}{(x+3)(x+5)}+\frac{2}{x+3} &=\frac{7}{(x+3)(x+5)}+\frac{2}{x+3} * \frac{x+5}{x+5} \\
    &=\frac{7}{(x+3)(x+5)}+\frac{2(x+5)}{(x+3)(x+5)} \\
    &=\frac{7+2 x+10}{(x+3)(x+5)} \\
    &=\frac{2 x+17}{(x+3)(x+5)}
    \end{aligned}
    \]
    Sometimes, the answer we end up with is not in lowest terms:

    Example
    Add the fractions.
    \[
    \begin{aligned}
    \frac{x}{x+2}+\frac{8}{x^{2}+8 x+12} & \\
    \frac{x}{x+2}+\frac{8}{x^{2}+8 x+12} &=\frac{x}{x+2}+\frac{8}{(x+2)(x+6)} \\
    &=\frac{x}{x+2} * \frac{x+6}{x+6}+\frac{8}{(x+2)(x+6)} \\
    &=\frac{x(x+6)}{(x+2)(x+6)}+\frac{8}{(x+2)(x+6)}=\frac{x(x+6)+8}{(x+2)(x+6)}
    \end{aligned}
    \]

    \[
    \frac{x(x+6)+8}{(x+2)(x+6)}=\frac{x^{2}+6 x+8}{(x+2)(x+6)}
    \]
    The numerator is factorable:
    \[
    \begin{aligned}
    \frac{x^{2}+6 x+8}{(x+2)(x+6)} &=\frac{(x+2)(x+4)}{(x+2)(x+6)} \\
    &=\frac{\cancel{(x+2)}(x+4)}{\cancel{(x+2)}(x+6)} \\
    &=\frac{x+4}{x+6}
    \end{aligned}
    \]

    Add or subtract the given expressions.
    1) \(\quad \frac{1}{x-1}-\frac{1}{x}\)
    2) \(\quad \frac{3}{y-6}-\frac{1}{y}\)
    3) \(\quad \frac{2}{x-3}+\frac{4}{x+3}\)
    4) \(\quad \frac{3}{x+4}-\frac{4}{x-2}\)
    5) \(\quad \frac{3}{k+2}-\frac{k-4}{k+5}\)
    6) \(\quad \frac{a+1}{a}-\frac{a}{a+1}\)
    7) \(\quad \frac{2 y}{y^{2}-25}-\frac{y}{y-5}\)
    8) \(\quad \frac{x}{x^{2}-1}+\frac{4}{x+1}\)
    9) \(\quad \frac{1}{x-3}+\frac{x}{x+1}\)
    10) \(\quad \frac{9 y}{y-4}-\frac{y+1}{y+5}\)
    Add or subtract the given expressions. Express your answers in lowest terms.
    11) \(\quad \frac{b}{b+1}-\frac{b+1}{2 b+2}\)
    12) \(\quad \frac{4 x+1}{8 x-12}+\frac{x-3}{2 x-3}\)
    13) \(\quad \frac{2}{a^{2}+4 a+3}+\frac{1}{a+3}\)
    14) \(\quad \frac{1}{y+6}-\frac{4}{y^{2}+8 y+12}\)
    15) \(\quad \frac{x+1}{2 x+4}-\frac{x^{2}}{2 x^{2}-8}\)
    16) \(\quad \frac{x+1}{x+2}-\frac{x^{2}+1}{x^{2}-x-6}\)
    17) \(\quad \frac{2 x}{x^{2}-3 x+2}+\frac{2 x}{x-1}-\frac{x}{x-2}\)
    18) \(\quad \frac{3 x+3}{2 x^{2}-x-1}+\frac{1}{2 x+1}\)
    19) \(\quad \frac{4 a}{a-2}-\frac{3 a}{a-3}+\frac{4 a}{a^{2}-5 a+6}\)
    20) \(\quad \frac{2}{y-3}-\frac{8-4 y}{y^{2}-8 y+15}\)
    21) \(\quad \frac{2 x}{x-1}+\frac{3 x}{x+1}-\frac{x+3}{x^{2}-1}\)
    22) \(\quad \frac{a}{a-1}-\frac{2}{a+2}+\frac{3(a-2)}{a^{2}+a-2}\)
    23) \(\quad \frac{x}{x-1}+\frac{x+7}{x^{2}-1}-\frac{x-2}{x+1}\)
    24) \(\quad \frac{2 y+5}{y^{2}-16}-\frac{y-9}{y^{2}-y-12}\)


    1.7: Adding and Subtracting Rational Expressions is shared under a not declared license and was authored, remixed, and/or curated by LibreTexts.

    • Was this article helpful?