Now solving each these linear equations gives,

\[\begin{align*}1 + 3u & = - 6u & \hspace{0.25in} & {\mbox{or}} & \hspace{0.25in}1 + 3u & = 6u\\ 1 & = - 9u & \hspace{0.25in} & {\mbox{or}} & \hspace{0.25in}1 & = 3u\\ u & = - \frac{1}{9} & \hspace{0.25in} & {\mbox{or}} & \hspace{0.25in}u & = \frac{1}{3}\end{align*}\] Show Step 3

Now, because the quantity outside of the absolute value bars was not a positive constant we need to be careful with the answers we got in the previous step. It is possible that one or both are not in fact solutions to the original equation. So, we need to verify each of the possible solutions from the previous step by checking them in the original equation.

\[u = - \frac{1}{9}:\,\,\,\,\,\,\,\,6\left( { - \frac{1}{9}} \right)\mathop = \limits^? \left| {1 + 3\left( { - \frac{1}{9}} \right)} \right|\,\,\,\,\,\,\,\, \to \,\,\,\,\,\,\, - \frac{2}{3}\mathop = \limits^? \left| {\frac{2}{3}} \right|\,\,\,\,\,\,\, \to \,\,\,\,\,\,\, - \frac{2}{3} \ne \frac{2}{3}\hspace{0.25in}{\mbox{NOT}}{\mbox{ OK}}\] \[u = \frac{1}{3}:\,\,\,\,\,\,\,\,6\left( {\frac{1}{3}} \right)\mathop = \limits^? \left| {1 + 3\left( {\frac{1}{3}} \right)} \right|\,\,\,\,\,\,\,\, \to \,\,\,\,\,\,\,2\mathop = \limits^? \left| 2 \right|\,\,\,\,\,\,\, \to \,\,\,\,\,\,\,2 = 2\hspace{0.25in}{\mbox{OK}}\]

Therefore, the only solution to the original equation is then : \(\require{bbox} \bbox[2pt,border:1px solid black]{{u = \frac{1}{3}}}\) .