Example 3
Completion requirements
| Example 3 |
Solve the rational equation \(\frac{{3k - 2}}{{k - 3}} - \frac{{k + 1}}{{k + 3}} = \frac{{k^2 + 5k - 6}}{{k^2 - 9}}\). Identify any non-permissible values, and verify the solution(s).
Step 1: Factor all polynomials, and identify the NPVs.
Step 2: Multiply each term by the LCD.
LCD: \((k - 3)(k + 3)\)
Step 3: Check for extraneous roots.
Recall that one of the NPVs is \(k \ne -3\); therefore, \(k = -3\) is an extraneous root.
Step 4: Verify for \(k = -1\).
The solution to the equation is \(k = -1\).
\[\begin{align}
\frac{{3k - 2}}{{k - 3}} - \frac{{k + 1}}{{k + 3}} &= \frac{{k^2 + 5k - 6}}{{k^2 - 9}} \\
\frac{{3k - 2}}{{k - 3}} - \frac{{k + 1}}{{k + 3}} &= \frac{{\left( {k + 6} \right)\left( {k - 1} \right)}}{{\left( {k + 3} \right)\left( {k - 3} \right)}} \\
\end{align}\]
\frac{{3k - 2}}{{k - 3}} - \frac{{k + 1}}{{k + 3}} &= \frac{{k^2 + 5k - 6}}{{k^2 - 9}} \\
\frac{{3k - 2}}{{k - 3}} - \frac{{k + 1}}{{k + 3}} &= \frac{{\left( {k + 6} \right)\left( {k - 1} \right)}}{{\left( {k + 3} \right)\left( {k - 3} \right)}} \\
\end{align}\]
\[\begin{align}
k - 3 &\ne 0 \\
k &\ne 3 \\
\end{align}\]
k - 3 &\ne 0 \\
k &\ne 3 \\
\end{align}\]
\[\begin{align}
k + 3 &\ne 0 \\
k &\ne -3 \\
\end{align}\]
k + 3 &\ne 0 \\
k &\ne -3 \\
\end{align}\]
Step 2: Multiply each term by the LCD.
LCD: \((k - 3)(k + 3)\)
\[\begin{align}
\left[ {\frac{{3k - 2}}{{{\color{red}\cancel{\color{#444}{k - 3}}}}}} \right]\left( {k + 3} \right){\color{red}\cancel {\color{#444}{\left( {k - 3} \right)}}} - \left[ {\frac{{k + 1}}{{{\color{red}\cancel{\color{#444}{k + 3}}}}}} \right]{\color{red}\cancel{\color{#444}{\left( {k + 3} \right)}}}\left( {k - 3} \right) &= \left[ {\frac{{\left( {k + 6} \right)\left( {k - 1} \right)}}{{{\color{red}\cancel{\color{#444}{\left( {k + 3} \right)\left( {k - 3} \right)}}}}}} \right]{\color{red}\cancel{\color{#444}{\left( {k + 3} \right)\left( {k - 3} \right)}}} \\
\left( {3k - 2} \right)\left( {k + 3} \right) - \left( {k + 1} \right)\left( {k - 3} \right) &= \left( {k + 6} \right)\left( {k - 1} \right) \\
3k^2 + 7k - 6 - \left( {k^2 - 2k - 3} \right) &= k^2 + 5k - 6 \\
2k^2 + 9k - 3 &= k^2 + 5k - 6 \\
k^2 + 4k + 3 &= 0 \\
\left( {k + 3} \right)\left( {k + 1} \right) &= 0 \\
k &= - 3 \thinspace {\rm{and}} \thinspace k = - 1 \\
\end{align}\]
\left[ {\frac{{3k - 2}}{{{\color{red}\cancel{\color{#444}{k - 3}}}}}} \right]\left( {k + 3} \right){\color{red}\cancel {\color{#444}{\left( {k - 3} \right)}}} - \left[ {\frac{{k + 1}}{{{\color{red}\cancel{\color{#444}{k + 3}}}}}} \right]{\color{red}\cancel{\color{#444}{\left( {k + 3} \right)}}}\left( {k - 3} \right) &= \left[ {\frac{{\left( {k + 6} \right)\left( {k - 1} \right)}}{{{\color{red}\cancel{\color{#444}{\left( {k + 3} \right)\left( {k - 3} \right)}}}}}} \right]{\color{red}\cancel{\color{#444}{\left( {k + 3} \right)\left( {k - 3} \right)}}} \\
\left( {3k - 2} \right)\left( {k + 3} \right) - \left( {k + 1} \right)\left( {k - 3} \right) &= \left( {k + 6} \right)\left( {k - 1} \right) \\
3k^2 + 7k - 6 - \left( {k^2 - 2k - 3} \right) &= k^2 + 5k - 6 \\
2k^2 + 9k - 3 &= k^2 + 5k - 6 \\
k^2 + 4k + 3 &= 0 \\
\left( {k + 3} \right)\left( {k + 1} \right) &= 0 \\
k &= - 3 \thinspace {\rm{and}} \thinspace k = - 1 \\
\end{align}\]
Step 3: Check for extraneous roots.
Recall that one of the NPVs is \(k \ne -3\); therefore, \(k = -3\) is an extraneous root.
Step 4: Verify for \(k = -1\).
| Left Side | Right Side |
|---|---|
|
\[\begin{array}{r} \frac{{3k - 2}}{{k - 3}} - \frac{{k + 1}}{{k + 3}} \\ \frac{{3\left( { - 1} \right) - 2}}{{\left( { - 1} \right) - 3}} - \frac{{\left( { - 1} \right) + 1}}{{\left( { - 1} \right) + 3}} \\ \frac{{ - 3 - 2}}{{ - 4}} - \frac{0}{2} \\ \frac{5}{4} \\ \end{array}\] |
\[\begin{array}{l} \frac{{k^2 + 5k - 6}}{{k^2 - 9}} \\ \frac{{\left( { - 1} \right)^2 + 5\left( { - 1} \right) - 6}}{{\left( { - 1} \right)^2 - 9}} \\ \frac{{1 - 5 - 6}}{{1 - 9}} \\ \frac{{ - 10}}{{ - 8}} \\ \frac{5}{4} \\ \end{array}\] |
| LS = RS | |
The solution to the equation is \(k = -1\).