Example  3
Simplify the rational expression \(\frac{{4\left( {x + 1} \right)}}{{6\left( {x + 3} \right)}} \cdot \frac{{3\left( {x + 2} \right)}}{{2\left( {x + 1} \right)}}\). Identify any non-permissible values.


Step 1: Identify the NPVs.

\(\begin{align}
 x + 3 &\ne 0 \\
 x &\ne -3 \\
 \end{align}\)
\(\begin{align}
 x + 1 &\ne 0 \\
 x &\ne -1 \\
 \end{align}\)

Step 2: Simplify the expression.

\[\begin{align}
 \frac{{{\color{red}\cancel {\color{#444}{4}^2}} {\color{red}\cancel {\color{#444}{\left(x + 1 \right)}}}}}{{{\color{red}\cancel {\color{#444}{6}^3}} \left( {x + 3} \right)}} \cdot \frac{{3\left( {x + 2} \right)}}{{2 {\color{red}\cancel {\color{#444}{\left(x + 1 \right)}}}}} &= \frac{{{\color{red}\cancel {\color{#444}{2}}}}}{{{\color{red}\cancel {\color{#444}{3}}} \left( {x + 3} \right)}} \cdot \frac{{{\color{red}\cancel {\color{#444}{3}}} \left( {x + 2} \right)}}{{{\color{red}\cancel {\color{#444}{2}}}}} \\
  &= \frac{1}{{x + 3}}\frac{{x + 2}}{1} \\
  &= \frac{{x + 2}}{{x + 3}},\thinspace x \ne - 3, - 1 \\
 \end{align}\]

Note that even though the \(x + 1\) term has been eliminated from the denominator, \(x = -1\) is still a non-permissible value.