Section 2.4 : Equations With More Than One Variable
In this section we are going to take a look at a topic that often doesn’t get the coverage that it deserves in an Algebra class. This is probably because it isn’t used in more than a couple of sections in an Algebra class. However, this is a topic that can, and often is, used extensively in other classes.
What we’ll be doing here is solving equations that have more than one variable in them. The process that we’ll be going through here is very similar to solving linear equations, which is one of the reasons why this is being introduced at this point. There is however one exception to that. Sometimes, as we will see, the ordering of the process will be different for some problems. Here is the process in the standard order.
- Multiply both sides by the LCD to clear out any fractions.
- Simplify both sides as much as possible. This will often mean clearing out parenthesis, canceling like terms, etc..
- Move all terms containing the variable we’re solving for to one side and all terms that don’t contain the variable to the opposite side.
- Get a single instance of the variable we’re solving for in the equation. For the types of problems that we’ll be looking at here this will almost always be accomplished by simply factoring the variable out of each of the terms.
- Divide by the coefficient of the variable. This step will make sense as we work problems. Note as well that in these problems the “coefficient” will probably contain things other than numbers.
It is usually easiest to see just what we’re going to be working with and just how they work with an example. We will also give the basic process for solving these inside the first example.
What we’re looking for here is an expression in the form,
\[r = \underline {{\mbox{Equation involving numbers, }}A,\,P,\,{\mbox{and }}t} \]In other words, the only place that we want to see an \(r\) is on the left side of the equal sign all by itself. There should be no other \(r\)’s anywhere in the equation. The process given above should do that for us.
Okay, let’s do this problem. We don’t have any fractions so we don’t need to worry about that. To simplify we will multiply the \(P\) through the parenthesis. Doing this gives,
\[A = P + Prt\]Now, we need to get all the terms with an \(r\) on them on one side. This equation already has that set up for us which is nice. Next, we need to get all terms that don’t have an \(r\) in them to the other side. This means subtracting a \(P\) from both sides.
\[A - P = Prt\]As a final step we will divide both sides by the coefficient of \(r\). Also, as noted in the process listed above the “coefficient” is not a number. In this case it is Pt. At this stage the coefficient of a variable is simply all the stuff that multiplies the variable.
\[\frac{{A - P}}{{Pt}} = r\hspace{0.5in} \Rightarrow \hspace{0.5in}r = \frac{{A - P}}{{Pt}}\]To get a final answer we went ahead and flipped the order to get the answer into a more “standard” form.
We will work more examples in a bit. However, let’s note a couple things first. These problems tend to seem fairly difficult at first, but if you think about it all we really did was use exactly the same process we used to solve linear equations. The main difference of course, is that there is more “mess” in this process. That brings us to the second point. Do not get excited about the mess in these problems. The problems will, on occasion, be a little messy, but the steps involved are steps that you can do! Finally, the answer will not be a simple number, but again it will be a little messy, often messier than the original equation. That is okay and expected.
Let’s work some more examples.
This one is fairly similar to the first example. However, it does work a little differently. Recall from the first example that we made the comment that sometimes the ordering of the steps in the process needs to be changed? Well, that’s what we’re going to do here.
The first step in the process tells us to clear fractions. However, since the fraction is inside a set of parentheses let’s first multiply the \(m\) through the parenthesis. Notice as well that if we multiply the \(m\) through first we will in fact clear one of the fractions out automatically. This will make our work a little easier when we do clear the fractions out.
\[V = \frac{m}{b} - 5aR\]Now, clear fractions by multiplying both sides by \(b\). We’ll also go ahead move all terms that don’t have an \(R\) in them to the other side.
\[\begin{align*}Vb & = m - 5abR\\ Vb - m & = - 5abR\end{align*}\]Be careful to not lose the minus sign in front of the 5! It’s very easy to lose track of that. The final step is to then divide both sides by the coefficient of the \(R\), in this case -5ab.
\[R = \frac{{Vb - m}}{{ - 5ab}} = - \frac{{Vb - m}}{{5ab}} = \frac{{ - \left( {Vb - m} \right)}}{{5ab}} = \frac{{ - Vb + m}}{{5ab}} = \frac{{m - Vb}}{{5ab}}\]Notice as well that we did some manipulation of the minus sign that was in the denominator so that we could simplify the answer somewhat.
In the previous example we solved for \(R\), but there is no reason for not solving for one of the other variables in the problems. For instance, consider the following example.
The first couple of steps are identical to the previous example. First, we will multiply the \(m\) through the parenthesis and then we will multiply both sides by \(b\) to clear the fractions. We’ve already done this work so from the previous example we have,
\[Vb - m = - 5abR\]In this case we’ve got \(b\)’s on both sides of the equal sign and we need all terms with \(b\)’s in them on one side of the equation and all other terms on the other side of the equation. In this case we can eliminate the minus signs if we collect the \(b\)’s on the left side and the other terms on the right side. Doing this gives,
\[Vb + 5abR = m\]Now, both terms on the right side have a \(b\) in them so if we factor that out of both terms we arrive at,
\[b\left( {V + 5aR} \right) = m\]Finally, divide by the coefficient of \(b\). Recall as well that the “coefficient” is all the stuff that multiplies the \(b\). Doing this gives,
\[b = \frac{m}{{V + 5aR}}\]First, multiply by the LCD, which is \(abc\) for this problem.
\[\begin{align*}\frac{1}{a}\left( {abc} \right) & = \left( {\frac{1}{b} + \frac{1}{c}} \right)\left( {abc} \right)\\ bc & = ac + ab\end{align*}\]Next, collect all the \(c\)’s on one side (the left will probably be easiest here), factor a \(c\) out of the terms and divide by the coefficient.
\[\begin{align*}bc - ac & = ab\\ c\left( {b - a} \right) & = ab\\ c & = \frac{{ab}}{{b - a}}\end{align*}\]First, we’ll need to clear the denominator. To do this we will multiply both sides by \(5x - 9\). We’ll also clear out any parenthesis in the problem after we do the multiplication.
\[\begin{align*}y\left( {5x - 9} \right) & = 4\\ 5xy - 9y & = 4\end{align*}\]Now, we want to solve for \(x\) so that means that we need to get all terms without a \(x\) in them to the other side. So, add 9\(y\) to both sides and the divide by the coefficient of \(x\).
\[\begin{align*}5xy & = 9y + 4\\ x & = \frac{{9y + 4}}{{5y}}\end{align*}\]This one is very similar to the previous example. Here is the work for this problem.
\[\begin{align*}y\left( {1 + 8x} \right) & = 4 - 3x\\ y + 8xy & = 4 - 3x\\ 8xy + 3x & = 4 - y\\ x\left( {8y + 3} \right) & = 4 - y\\ x & = \frac{{4 - y}}{{8y + 3}}\end{align*}\]As mentioned at the start of this section we won’t be seeing this kind of problem all that often in this class. However, outside of this class (a Calculus class for example) this kind of problem shows up with surprising regularity.