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### Section 2.2 : Linear Equations

$\frac{{4 - 2z}}{3} = \frac{3}{4} - \frac{{5z}}{6}$

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The first step here is to multiply both sides by the LCD, which happens to be 12 for this problem.

\begin{align*}12\left( {\frac{{4 - 2z}}{3}} \right) & = 12\left( {\frac{3}{4} - \frac{{5z}}{6}} \right)\\ 12\left( {\frac{{4 - 2z}}{3}} \right) & = 12\left( {\frac{3}{4}} \right) - 12\left( {\frac{{5z}}{6}} \right)\\ 4\left( {4 - 2z} \right) & = 3\left( 3 \right) - 2\left( {5z} \right)\end{align*} Show Step 2

Now we need to find the solution and so all we need to do is go through the same process that we used in the first two practice problems. Here is that work.

\begin{align*}4\left( {4 - 2z} \right) & = 3\left( 3 \right) - 2\left( {5z} \right)\\ 16 - 8z & = 9 - 10z\\ 2z & = - 7\\ z & = - \frac{7}{2}\end{align*} Show Step 3

Now all we need to do is check our answer from Step 2 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 2 into the equation given in the problem statement.

Here is the verification work.

\begin{align*}\frac{{4 - 2\left( { - \frac{7}{2}} \right)}}{3} & \mathop = \limits^? \frac{3}{4} - \frac{{5\left( { - \frac{7}{2}} \right)}}{6}\\ \frac{{4 + 7}}{3} & \mathop = \limits^? \frac{3}{4} - \frac{{ - \frac{{35}}{2}}}{6}\\ \frac{{11}}{3} & \mathop = \limits^? \frac{3}{4} + \frac{{35}}{{12}}\\ \frac{{11}}{3} & = \frac{{11}}{3}\hspace{0.2in} {\mbox{OK}}\end{align*}

So, we can see that our solution from Step 2 is in fact the solution to the equation.

Note that the verification work can often be quite messy so don’t get excited about it when it does. Verification is an important step to always remember for these kinds of problems. You should always know if you got the answer correct before you check the answers and/or your instructor grades the problem!