Example  2
Solve the rational equation \(\frac{2}{{x^2  - 16}} + \frac{1}{{x + 4}} = \frac{3}{{x - 4}}\). Identify any non-permissible values, and verify the solution(s).


Step 1: Factor all polynomials to determine the non-permissible values and the lowest common denominator.

\[\begin{align}
 \frac{2}{{x^2 - 16}} + \frac{1}{{x + 4}} &= \frac{3}{{x - 4}} \\
 \frac{2}{{\left( {x - 4} \right)\left( {x + 4} \right)}} + \frac{1}{{x + 4}} &= \frac{3}{{x - 4}} \\
 \end{align}\]

NPVs:

\(\begin{align}
 x - 4 &\ne 0 \\
 x &\ne 4 \\
 \end{align}\)
\(\begin{align}
 x + 4 &\ne 0 \\
 x &\ne -4 \\
 \end{align}\)

LCD: \((x - 4)(x + 4)\)

Step 2: Multiply each term by the LCD.

\[\begin{align}
 \left[ {\frac{2}{{{\color{red}\cancel {\color{#444}{(x - 4)(x + 4)}}}}}} \right] {\color{red}\cancel {\color{#444}{(x - 4)(x + 4)}}} + \left[ {\frac{1}{{{\color{red}\cancel{\color{#444}{x + 4}}}}}} \right]\left( {x - 4} \right) {\color{red}\cancel {\color{#444}{\left ( x + 4 \right)}}} &= \left[ {\frac{3}{{{\color{red}\cancel{\color{#444}{x - 4}}}}}} \right]{\color{red}\cancel{\color{#444}{\left(x - 4\right)}}}\left( {x + 4} \right) \\
 2 + \left( {x - 4} \right) &= 3\left( {x + 4} \right) \\
 2 + x - 4 &= 3x + 12 \\
 x - 2 &= 3x + 12 \\
 \end{align}\]

Step 3: Solve for \(x\).

\(\begin{align}
 x - 2 &= 3x + 12 \\
  - 14 &= 2x \\
  - 7 &= x \\
 \end{align}\)

Step 4: Verify for \(x = -7\).

Left Side Right Side
\[\begin{array}{r}
 \frac{2}{{x^2 - 16}} + \frac{1}{{x + 4}} \\
 \frac{2}{{\left( { - 7} \right)^2 - 16}} + \frac{1}{{\left( { - 7} \right) + 4}} \\
 \frac{2}{{49 - 16}} + \frac{1}{{ - 3}} \\
 \frac{2}{{33}} - \frac{1}{3} \\
 \frac{2}{{33}} - \frac{{11}}{{33}} \\
 \frac{{ - 9}}{{33}} \\
  - \frac{3}{{11}} \\
 \end{array}\]
 \[\begin{array}{l}
 \frac{3}{{x - 4}} \\
 \frac{3}{{\left( { - 7} \right) - 4}} \\
 \frac{3}{{ - 11}} \\
  - \frac{3}{{11}} \\
 \end{array}\]
LS = RS \(\hspace{33pt}\)

Because the left side equals the right side, and \(x = -7\) is not an NPV, \(x = -7\) is the solution to the equation.