Example 4
Completion requirements
| Example 4 |
Solve the rational equation \(\frac{{2w - 2}}{{w^2 - 1}} = \frac{2}{{w + 2}}\). Identify any non-permissible values, and verify the solution(s).
Step 1: Factor all polynomials, and identify the NPVs.
Step 2: Simplify, and determine the LCD.
LCD: \((w + 1)(w + 2)\)
Step 3: Multiply each term by the LCD and solve.
The result is a false statement, which tells you that there are no values for the variable that satisfy the equation. Also note that the variable was eliminated while solving this equation. As a result, there is no solution to this equation.
\[\begin{align}
\frac{{2w - 2}}{{w^2 - 1}} &= \frac{2}{{w + 2}} \\
\frac{{2\left( {w - 1} \right)}}{{\left( {w - 1} \right)\left( {w + 1} \right)}} &= \frac{2}{{w + 2}} \\
\end{align}\]
\frac{{2w - 2}}{{w^2 - 1}} &= \frac{2}{{w + 2}} \\
\frac{{2\left( {w - 1} \right)}}{{\left( {w - 1} \right)\left( {w + 1} \right)}} &= \frac{2}{{w + 2}} \\
\end{align}\]
\(\begin{align}
w - 1 &\ne 0 \\
w &\ne 1 \\
\end{align}\)
w - 1 &\ne 0 \\
w &\ne 1 \\
\end{align}\)
\(\begin{align}
w + 1 &\ne 0 \\
w &\ne -1 \\
\end{align}\)
w + 1 &\ne 0 \\
w &\ne -1 \\
\end{align}\)
\(\begin{align}
w + 2 &\ne 0 \\
w &\ne -2 \\
\end{align}\)
w + 2 &\ne 0 \\
w &\ne -2 \\
\end{align}\)
Step 2: Simplify, and determine the LCD.
\[\begin{align}
\frac{{2{\color{red}\cancel{\color{#444} {\left(w - 1\right)}}}}}{{{\color{red}\cancel {\color{#444}{\left(w - 1\right)}}}\left( {w + 1} \right)}} &= \frac{2}{{w + 2}} \\
\frac{2}{{w + 1}} &= \frac{2}{{w + 2}} \\
\end{align}\]
\frac{{2{\color{red}\cancel{\color{#444} {\left(w - 1\right)}}}}}{{{\color{red}\cancel {\color{#444}{\left(w - 1\right)}}}\left( {w + 1} \right)}} &= \frac{2}{{w + 2}} \\
\frac{2}{{w + 1}} &= \frac{2}{{w + 2}} \\
\end{align}\]
LCD: \((w + 1)(w + 2)\)
Step 3: Multiply each term by the LCD and solve.
\[\begin{align}
\left[ {\frac{2}{{{\color{red}\cancel{\color{#444}{w + 1}}}}}} \right]{\color{red}\cancel{\color{#444}{\left( {w + 1} \right)}}}\left( {w + 2} \right) &= \left[ {\frac{2}{{{\color{red}\cancel{\color{#444}{w + 2}}}}}} \right]\left( {w + 1} \right) {\color{red}\cancel{\color{#444}{\left( {w + 2} \right)}}}\\
2\left( {w + 2} \right) &= 2\left( {w + 1} \right) \\
2w + 4 &= 2w + 2 \\
4 &= 2 \\
\end{align}\]
\left[ {\frac{2}{{{\color{red}\cancel{\color{#444}{w + 1}}}}}} \right]{\color{red}\cancel{\color{#444}{\left( {w + 1} \right)}}}\left( {w + 2} \right) &= \left[ {\frac{2}{{{\color{red}\cancel{\color{#444}{w + 2}}}}}} \right]\left( {w + 1} \right) {\color{red}\cancel{\color{#444}{\left( {w + 2} \right)}}}\\
2\left( {w + 2} \right) &= 2\left( {w + 1} \right) \\
2w + 4 &= 2w + 2 \\
4 &= 2 \\
\end{align}\]
The result is a false statement, which tells you that there are no values for the variable that satisfy the equation. Also note that the variable was eliminated while solving this equation. As a result, there is no solution to this equation.