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Solve and verify the rational equations.

  1. \(\frac{4}{{2x - 1}} = \frac{1}{{x - 2}}\)

    NPVs:

    \[\begin{align}
     2x - 1 &\ne 0 \\
     x &\ne \frac{1}{2} \\
     \end{align}\]
    \[\begin{align}
     x - 2 &\ne 0 \\
     x &\ne 2 \\
     \end{align}\]

    \[\begin{align}
     \left[ {\frac{4}{{{\color{red}\cancel{\color{#444}{2x - 1}}}}}} \right]{\color{red}\cancel {\color{#444}{(2x - 1)}}}\left( {x - 2} \right) &= \left[ {\frac{1}{{{\color{red}\cancel{\color{#444}{x - 2}}}}}} \right]\left( {2x - 1} \right) {\color{red}\cancel{\color{#444}{(x - 2)}}}\\
     4\left( {x - 2} \right) &= 2x - 1 \\
     4x - 8 &= 2x - 1 \\
     2x &= 7 \\
     x &= \frac{7}{2} \\
     \end{align}\]


    Verify for \(x = \frac{7}{2}\).

    Left Side Right Side
    \[\begin{array}{r}
     \frac{4}{{2x - 1}} \\
     \frac{4}{{2\left( {\frac{7}{2}} \right) - 1}} \\
     \frac{4}{{7 - 1}} \\
     \frac{4}{6} \\
     \frac{2}{3} \\
     \end{array}\]
    		\[\begin{array}{l}
     \frac{1}{{x - 2}} \\
     \frac{1}{{\left( {\frac{7}{2}} \right) - 2}} \\
     \frac{1}{{\frac{{7 - 4}}{2}}} \\
     1 \cdot \frac{2}{3} \\
     \frac{2}{3} \\
     \end{array}\]
    LS = RS

    The solution to the equation is \(x = \frac{7}{2}\).

  2. \(\frac{{12}}{{10 + y}} + \frac{{12}}{{10 - y}} = \frac{5}{2}\)

    NPVs:

    \(\begin{align}
     10 + y &\ne 0 \\
     y &\ne -10 \\
     \end{align}\)
    \(\begin{align}
     10 - y &\ne 0 \\
     y &\ne 10 \\
     \end{align}\)

    \[\begin{align}
     \left[ {\frac{{12}}{{{\color{red}\cancel{\color{#444}{10 + y}}}}}} \right]2 {\color{red}\cancel{\color{#444}{\left( {10 + y} \right)}}}\left( {10 - y} \right) + \left[ {\frac{{12}}{{{\color{red}\cancel{\color{#444}{10 - y}}}}}} \right]2\left( {10 + y} \right){\color{red}\cancel{\color{#444}{\left( {10 - y} \right)} }}&= \left[ {\frac{5}{{{\color{red}\cancel{\color{#444}{2}}}}}} \right]{\color{red}\cancel{\color{#444}{2}}}\left( {10 + y} \right)\left( {10 - y} \right) \\
     12\left( 2 \right)\left( {10 - y} \right) + 12\left( 2 \right)\left( {10 + y} \right) &= 5\left( {10 + y} \right)\left( {10 - y} \right) \\
     240 - 24y + 240 + 24y &= 5\left( {100 - y^2 } \right) \\
     480 &= 5\left( {100 - y^2 } \right) \\
     96 &= 100 - y^2 \\
     y^2 &= 4 \\
     y &= \pm 2 \\
     \end{align}\]

    Verify for \(y = 2\).

    Left Side Right Side
    \[\begin{array}{r}
     \frac{{12}}{{10 + y}} + \frac{{12}}{{10 - y}} \\
     \frac{{12}}{{10 + \left( 2 \right)}} + \frac{{12}}{{10 - \left( 2 \right)}} \\
     \frac{{12}}{{12}} + \frac{{12}}{8} \\
     1 + \frac{3}{2} \\
     \frac{5}{2} \\
     \end{array}\]
    		 \[\frac{5}{2}\]
    LS = RS \(\hspace{30pt}\)

    Verify for \(y = -2\).

    Left Side Right Side
    \[\begin{array}{r}
     \frac{{12}}{{10 + y}} + \frac{{12}}{{10 - y}} \\
     \frac{{12}}{{10 + \left( { - 2} \right)}} + \frac{{12}}{{10 - \left( { - 2} \right)}} \\
     \frac{{12}}{8} + \frac{{12}}{{12}} \\
     \frac{3}{2} + 1 \\
     \frac{5}{2} \\
     \end{array}\]
    		 \[\frac{5}{2}\]
    LS = RS \(\hspace{30pt}\)

    The solution to the equation is \(y = \pm 2\).