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Section 2.2 : Linear Equations

5. Solve the following equation and check your answer.

\[\frac{{3y + 4}}{{y - 1}} = 2 + \frac{7}{{y - 1}}\]

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Hint : Do not forget to watch out for values of \(y\) that we’ll need to avoid!
Start Solution

First, we can see that the LCD for this equation is,

\[y - 1\]

From this we can also see that we’ll need to avoid \(y = 1\). Remember that we have to avoid division by zero and we will clearly get division by zero with this value of \(y\).

Show Step 2

Next, we need to do find the solution. To get the solution we’ll need to multiply both sides by the LCD and the go through the same process we used in the first couple of practice problems. Here is that work.

\[\begin{align*}\left( {y - 1} \right)\left( {\frac{{3y + 4}}{{y - 1}}} \right) & = \left( {2 + \frac{7}{{y - 1}}} \right)\left( {y - 1} \right)\\ 3y + 4 & = 2\left( {y - 1} \right) + 7\\ 3y + 4 & = 2y + 5\\ y & = 1\end{align*}\] Show Step 3

Finally, we need to verify that our answer from Step 2 is in fact a solution and in this case there isn’t a lot of work to that process. We can see that our potential solution from Step 2 is in fact the value of \(y\) that we need to avoid and so this equation has no solution.

We could also see this if we plugged the value of \(y\) from Step 2 into the equation given in the problem statement. Had we done that we would have gotten a division by zero in two of the terms! That, of course, is why we needed to avoid \(y = 1\) .

Note as well that we only have caught the division by zero if we verify by plugging into the equation in the problem statement. Had we checked in the equation we got by multiplying by the LCD it would have appeared to be a solution! This is the reason that we need to always check in the equation from the problem statement.