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Section 2.2 : Linear Equations

2. Solve the following equation and check your answer.

$2\left( {w + 3} \right) - 10 = 6\left( {32 - 3w} \right)$

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Start Solution

First, we need to clear out the parenthesis on each side and then simplify each side.

\begin{align*}2\left( {w + 3} \right) - 10 & = 6\left( {32 - 3w} \right)\\ 2w + 6 - 10 & = 192 - 18w\\ 2w - 4 & = 192 - 18w\end{align*} Show Step 2

Now we can add 18$$w$$ and 4 to both sides to get all the $$w$$’s on one side and the terms without an $$w$$ on the other side.

\begin{align*}2w - 4 & = 192 - 18w\\ 20w & = 196\end{align*} Show Step 3

Finally, all we need to do is divide both sides by the coefficient of the $$w$$ (i.e. the 20) to get the solution of $$w = \frac{{196}}{{20}} = \frac{{49}}{5}$$.

Don’t get excited about solutions that are fractions. They happen more often than people tend to realize.

Show Step 4

Now all we need to do is check our answer from Step 3 and verify that it is a solution to the equation. It is important when doing this step to verify by plugging the solution from Step 3 into the equation given in the problem statement.

Here is the verification work.

\begin{align*}2\left( {\frac{{49}}{5} + 3} \right) - 10 & \mathop = \limits^? 6\left( {32 - 3\left( {\frac{{49}}{5}} \right)} \right)\\ 2\left( {\frac{{64}}{5}} \right) - 10 & \mathop = \limits^? 6\left( {\frac{{13}}{5}} \right)\\ \frac{{78}}{5} & = \frac{{78}}{5}\hspace{0.5in}{\mbox{OK}}\end{align*}

So, we can see that our solution from Step 3 is in fact the solution to the equation.