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Solve each rational equation, if possible, and identify the non-permissible values.

  1. \(\frac{3}{{t + 4}} = \frac{9}{{3t - 5}}\)

    NPVs:

    \[\begin{align}
     t + 4 &\ne 0 \\
     t &\ne - 4 \\
      \\
     \end{align}\]
    \[\begin{align}
     3t - 5 &\ne 0 \\
     t &\ne \frac{5}{3} \\
     \end{align}\]

    LCD: \((t + 4)(3t - 5)\)

    \[\begin{align}
     \left[ {\frac{3}{{{\color{red}\cancel{\color{#444}{t + 4}}}}}} \right]{\color{red}\cancel{\color{#444}{\left( {t + 4} \right)}}}\left( {3t - 5} \right) &= \left[ {\frac{9}{{{\color{red}\cancel{\color{#444}{3t - 5}}}}}} \right]\left( {t + 4} \right){\color{red}\cancel{\color{#444}{\left( {3t - 5} \right)}}} \\
     3\left( {3t - 5} \right) &= 9\left( {t + 4} \right) \\
     9t - 15 &= 9t + 36 \\
      - 15 &= 36 \\
     \end{align}\]

    The result is a false statement. There is no solution to this equation.

  2. \(\frac{{2x + 3}}{{x - 2}} + \frac{{5x - 1}}{{3x + 1}} = \frac{{7x^2 + 9x + 3}}{{3x^2 - 5x - 2}}\)

    \[\frac{{2x + 3}}{{x - 2}} + \frac{{5x - 1}}{{3x + 1}} = \frac{{7x^2 + 9x + 3}}{{\left( {3x + 1} \right)\left( {x - 2} \right)}} \]

    NPVs:

    \[\begin{align}
     x - 2 &\ne 0 \\
     x &\ne 2 \\
     \end{align}\]
    \[\begin{align}
     3x + 1 &\ne 0 \\
     x &\ne -\frac{1}{3} \\
     \end{align}\]

    LCD: \((x - 2)(3x + 1)\)

    \[\begin{align}
     \left[ {\frac{{2x + 3}}{{{\color{red}\cancel{\color{#444}{x - 2}}}}}} \right]\left( {3x + 1} \right) {\color{red}\cancel{\color{#444}{\left(x - 2\right)}}} + \left[ {\frac{{5x - 1}}{{{\color{red}\cancel{\color{#444}{3x + 1}}}}}} \right]{\color{red}\cancel{\color{#444}{\left(3x + 1\right)}}}\left( {x - 2} \right) &= \left[ {\frac{{7x^2 + 9x + 3}}{{{\color{red}\cancel{\color{#444}{\left(3x + 1\right) \left(x - 2\right)}}}}}} \right] {\color{red}\cancel{\color{#444}{\left(3x + 1\right) \left(x - 2\right)}}}\\
     \left( {2x + 3} \right)\left( {3x + 1} \right) + \left( {5x - 1} \right)\left( {x - 2} \right) &= 7x^2 + 9x + 3 \\
     6x^2 + 11x + 3 + 5x^2 - 11x + 2 &= 7x^2 + 9x + 3 \\
     11x^2 + 5 &= 7x^2 + 9x + 3 \\
     4x^2 - 9x + 2 &= 0 \\
     \left( {4x - 1} \right)\left( {x - 2} \right) &= 0 \\
     x &= \frac{1}{4}\thinspace {\rm{and}}\thinspace x = 2 \\
     \end{align}\]

    Note that \(x = 2\) is an extraneous root because it is a non-permissible value.

    Verify for \(x = \frac{1}{4}\).

    Left Side Right Side
    \[\begin{array}{r}
     \frac{{2x + 3}}{{x - 2}} + \frac{{5x - 1}}{{3x + 1}} \\
     \frac{{2\left( {\frac{1}{4}} \right) + 3}}{{\left( {\frac{1}{4}} \right) - 2}} + \frac{{5\left( {\frac{1}{4}} \right) - 1}}{{3\left( {\frac{1}{4}} \right) + 1}} \\
     \left[ {\frac{{\frac{1}{2} + 3}}{{ - \frac{7}{4}}}} \right] \cdot \left( {\frac{4}{4}} \right) + \left[ {\frac{{\frac{5}{4} - 1}}{{\frac{3}{4} + 1}}} \right] \cdot \left( {\frac{4}{4}} \right) \\
     \frac{{2 + 12}}{{ - 7}} + \frac{{5 - 4}}{{3 + 4}} \\
      - \frac{{14}}{7} + \frac{1}{7} \\
      - \frac{{13}}{7} \\
     \end{array}\]
    		\[\begin{array}{l}
     \frac{{7x^2 + 9x + 3}}{{3x^2 - 5x - 2}} \\
     \frac{{7\left( {\frac{1}{4}} \right)^2 + 9\left( {\frac{1}{4}} \right) + 3}}{{3\left( {\frac{1}{4}} \right)^2 - 5\left( {\frac{1}{4}} \right) - 2}} \\
     \frac{{\left[ {\frac{7}{{16}} + \frac{9}{4} + 3} \right]}}{{\left[ {\frac{3}{{16}} - \frac{5}{4} - 2} \right]}}\cdot \left( {\frac{{16}}{{16}}} \right) \\
     \frac{{7 + 36 + 48}}{{3 - 20 - 32}} \\
     \frac{{91}}{{ - 49}} \\
      - \frac{{13}}{7} \\
     \end{array}\]
                                 LS = RS

    The solution to the equation is \(x = \frac{1}{4}\).


 For further information about solving rational equations see pp. 342 to 344 of Pre-Calculus 11.