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In the following problems we will describe in detail the first problem and the leave most of the explanation out of the following problems.

a \(3\left( {x + 5} \right) = 2\left( { - 6 - x} \right) - 2x\)

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For this problem there are no fractions so we don’t need to worry about the first step in the process. The next step tells to simplify both sides. So, we will clear out any parenthesis by multiplying the numbers through and then combine like terms.

\[\begin{align*}3\left( {x + 5} \right) & = 2\left( { - 6 - x} \right) - 2x\\ 3x + 15 & = - 12 - 2x - 2x\\ 3x + 15 & = - 12 - 4x\end{align*}\]

The next step is to get all the \(x\)’s on one side and all the numbers on the other side. Which side the \(x\)’s go on is up to you and will probably vary with the problem. As a rule of thumb, we will usually put the variables on the side that will give a positive coefficient. This is done simply because it is often easy to lose track of the minus sign on the coefficient and so if we make sure it is positive we won’t need to worry about it.

So, for our case this will mean adding 4\(x\) to both sides and subtracting 15 from both sides. Note as well that while we will actually put those operations in this time we normally do these operations in our head.

\[\begin{align*}\require{color} 3x + 15 & = - 12 - 4x\\ 3x + 15 {\color{Red} - 15}{\color{Blue} + 4x} & = - 12 - 4x {\color{Blue} + 4x} {\color{Red} - 15}\\ 7x & = - 27\end{align*}\]

The next step says to get a coefficient of 1 in front of the \(x\). In this case we can do this by dividing both sides by a 7.

\[\begin{align*}\frac{{7x}}{7} &= \frac{{ - 27}}{7}\\ x & = - \frac{{27}}{7}\end{align*}\]

Now, if we’ve done all of our work correct \(x = - \frac{{27}}{7}\) is the solution to the equation.

The last and final step is to then check the solution. As pointed out in the process outline we need to check the solution in the **original** equation. This is important, because we may have made a mistake in the very first step and if we did and then checked the answer in the results from that step it may seem to indicate that the solution is correct when the reality will be that we don’t have the correct answer because of the mistake that we originally made.

The problem of course is that, with this solution, that checking might be a little messy. Let’s do it anyway.

\[\begin{align*}3\left( { - \frac{{27}}{7} + 5} \right) & \mathop = \limits^? 2\left( { - 6 - \left( { - \frac{{27}}{7}} \right)} \right) - 2\left( { - \frac{{27}}{7}} \right)\\ 3\left( {\frac{8}{7}} \right) & \mathop = \limits^? 2\left( { - \frac{{15}}{7}} \right) + \frac{{54}}{7}\\ \frac{{24}}{7} & = \frac{{24}}{7}\hspace{0.5in}{\mbox{OK}}\end{align*}\]

So, we did our work correctly and the solution to the equation is,

\[x = - \frac{{27}}{7}\]

Note that we didn’t use the solution set notation here. For single solutions we will rarely do that in this class. However, if we had wanted to the solution set notation for this problem would be,

\[\left\{ { - \frac{{27}}{7}} \right\}\]

Before proceeding to the next problem let’s first make a quick comment about the “messiness’ of this answer. Do NOT expect all answers to be nice simple integers. While we do try to keep most answer simple often they won’t be so do NOT get so locked into the idea that an answer must be a simple integer that you immediately assume that you’ve made a mistake because of the “messiness” of the answer.

b \(\displaystyle \frac{{m - 2}}{3} + 1 = \frac{{2m}}{7}\)

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Okay, with this one we won’t be putting quite as much explanation into the problem.

In this case we have fractions so to make our life easier we will multiply both sides by the LCD, which is 21 in this case. After doing that the problem will be very similar to the previous problem. Note as well that the denominators are only numbers and so we won’t need to worry about division by zero issues.

Let’s first multiply both sides by the LCD.

\[\begin{align*}21\left( {\frac{{m - 2}}{3} + 1} \right) & = \left( {\frac{{2m}}{7}} \right)21\\ 21\left( {\frac{{m - 2}}{3}} \right) + 21\left( 1 \right) & = \left( {\frac{{2m}}{7}} \right)21\\ 7\left( {m - 2} \right) + 21 & = \left( {2m} \right)\left( 3 \right)\end{align*}\]

Be careful to correctly distribute the 21 through the parenthesis on the left side. Everything inside the parenthesis needs to be multiplied by the 21 before we simplify. At this point we’ve got a problem that is similar the previous problem and we won’t bother with all the explanation this time.

\[\begin{align*}7\left( {m - 2} \right) + 21 & = \left( {2m} \right)\left( 3 \right)\\ 7m - 14 + 21 & = 6m\\ 7m + 7 & = 6m\\ m & = - 7\end{align*}\]

So, it looks like \(m = - 7\) is the solution. Let’s verify it to make sure.

\[\begin{align*}\frac{{ - 7 - 2}}{3} + 1 & \mathop = \limits^? \frac{{2\left( { - 7} \right)}}{7}\\ \frac{{ - 9}}{3} + 1 & \mathop = \limits^? - \frac{{14}}{7}\\ - 3 + 1 &\mathop = \limits^? - 2\\ - 2 & = - 2\hspace{0.5in}{\mbox{OK}}\end{align*}\]

So, it is the solution.

c \(\displaystyle \frac{5}{{2y - 6}} = \frac{{10 - y}}{{{y^2} - 6y + 9}}\)

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This one is similar to the previous one except now we’ve got variables in the denominator. So, to get the LCD we’ll first need to completely factor the denominators of each rational expression.

\[\frac{5}{{2\left( {y - 3} \right)}} = \frac{{10 - y}}{{{{\left( {y - 3} \right)}^2}}}\]

So, it looks like the LCD is \(2{\left( {y - 3} \right)^2}\). Also note that we will need to avoid \(y = 3\) since if we plugged that into the equation we would get division by zero.

Now, outside of the \(y\)’s in the denominator this problem works identical to the previous one so let’s do the work.

\[\begin{align*}\left( 2 \right){\left( {y - 3} \right)^2}\left( {\frac{5}{{2\left( {y - 3} \right)}}} \right) & = \left( {\frac{{10 - y}}{{{{\left( {y - 3} \right)}^2}}}} \right)\left( 2 \right){\left( {y - 3} \right)^2}\\ 5\left( {y - 3} \right) & = 2\left( {10 - y} \right)\\ 5y - 15 & = 20 - 2y\\ 7y & = 35\\ & y = 5\end{align*}\]

Now the solution is not \(y = 3\) so we won’t get division by zero with the solution which is a good thing. Finally, let’s do a quick verification.

\[\begin{align*}\frac{5}{{2\left( 5 \right) - 6}} & \mathop = \limits^? \frac{{10 - 5}}{{{5^2} - 6\left( 5 \right) + 9}}\\ \frac{5}{4} & = \frac{5}{4}\hspace{0.5in}{\mbox{OK}}\end{align*}\]

d \(\displaystyle \frac{{2z}}{{z + 3}} = \frac{3}{{z - 10}} + 2\)

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In this case it looks like the LCD is \(\left( {z + 3} \right)\left( {z - 10} \right)\) and it also looks like we will need to avoid \(z = - 3\) and \(z = 10\) to make sure that we don’t get division by zero.

Let’s get started on the work for this problem.

\[\begin{align*}\left( {z + 3} \right)\left( {z - 10} \right)\left( {\frac{{2z}}{{z + 3}}} \right) & = \left( {\frac{3}{{z - 10}} + 2} \right)\left( {z + 3} \right)\left( {z - 10} \right)\\ 2z\left( {z - 10} \right) & = 3\left( {z + 3} \right) + 2\left( {z + 3} \right)\left( {z - 10} \right)\\ 2{z^2} - 20z & = 3z + 9 + 2\left( {{z^2} - 7z - 30} \right)\end{align*}\]

At this point let’s pause and acknowledge that we’ve got a *z*^{2} in the work here. Do not get excited about that. Sometimes these will show up temporarily in these problems. You should only worry about it if it is still there after we finish the simplification work.

So, let’s finish the problem.

\[\begin{align*}\require{cancel} \cancel{{2{z^2}}} - 20z & = 3z + 9 + \cancel{{2{z^2}}} - 14z - 60\\ - 20z & = - 11z - 51\\ 51 & = 9z\\ \frac{{51}}{9} & = z\\ & \frac{{17}}{3} = z\end{align*}\]

Notice that the *z*^{2} did in fact cancel out. Now, if we did our work correctly \(z = \frac{{17}}{3}\) should be the solution since it is not either of the two values that will give division by zero. Let’s verify this.

\[\begin{align*}\frac{{2\left( {\frac{{17}}{3}} \right)}}{{\frac{{17}}{3} + 3}} & \mathop = \limits^? \frac{3}{{\frac{{17}}{3} - 10}} + 2\\ \frac{{\,\,\frac{{34}}{3}\,\,}}{{\frac{{26}}{3}}} & \mathop = \limits^? \frac{3}{{ - \frac{{13}}{3}}} + 2\\ \frac{{34}}{3}\left( {\frac{3}{{26}}} \right) & \mathop = \limits^? 3\left( { - \frac{3}{{13}}} \right) + 2\\ \frac{{17}}{{13}} & = \frac{{17}}{{13}}\hspace{0.5in}{\mbox{OK}}\end{align*}\]

The checking can be a little messy at times, but it does mean that we KNOW the solution is correct.