(An archive question of the week)
Last time I discussed issues that arise in solving a simple algebraic equation. In researching that, I found a discussion of solving a formula for a variable (which in some countries is called “making x the subject”, that is, changing an equation involving x into the form “x = …”). This brings up several very common errors, and ways to help a beginning student get it right.
How do I solve a formula for a variable?
The question comes from 2006:
Manipulating a Formula Algebraically I have the formula M = C(1 + r) in my algebra text, but it doesn't tell me what it is for. It also asks me to solve for the r variable. I know that the formula when solved for the r variable is: M - C r = ----- C but I don't understand how to get all the way there. This is as far as I know: M = C(1 + r) -C -C ---------------- M - C = (1 + r) Beyond this I haven't got a clue. Where did the 1 go? Where does the extra C come from? How do you get r alone? Thanks for your help!
Note that Debbie knows the correct answer, so I am free to show how to get it. Otherwise, I probably would have offered a similar example to demonstrate what to do, and then let her do her own (hoping that I showed the right things). Here, this problem is the example. Showing us her work gave me both a clear picture of the type of help she needed (she made some classic errors), and a place to start in demonstrating correct methods and how to recognize mistakes.
What formula is it?
But first, I dealt with the question of what the formula is for (which was the subject line of her question). Often formulas are first introduced by showing a variety of practical formulas with brief explanations. But then when students are asked to solve for a variable, the book may not bother saying what the variables mean, or even what subject the formula arises in. In fact, the formula may be entirely made up. None of this really matters:
Hi, Debbie. First, you don't need to even ask what a formula is for in this sort of problem. One of the most important things about algebra (or math in general, really) is that it is abstract: that is, we can take all sorts of real-world problems and turn them into math problems (such as equations to solve), and once you've done that, it doesn't matter at all where they came from. The methods you use in algebra ignore the meaning of the problem, and just look at the equation itself. Many equations you'll work with in class don't come from anywhere at all; you are just practicing techniques that you can use on problems that do have some real-world meaning. It's sort of like a medical student practicing an operation on a dummy; he doesn't have to ask about the patient's family or insurance! But when he gets into real medicine, all those things will matter. In this case, the equation MIGHT relate to interest on a bank account, with r being the interest rate; but the same equation could come from other sorts of problems, too--or none at all.
The formula, \(M = C(1 + r)\), appears to be some sort of percentage increase (markup, tax, interest, …), but the letters M and C might stand for anything. (Here, I suspect they are something like Marked-up price and Cost.) In solving for r, in an applied problem we would be finding the percent increase; but this problem is only giving practice that will later be useful for such applications.
Fixing the error
Students often have trouble when they first see this kind of problem (sometimes called “literal equations” because they contain only letters, not numbers), in which, rather than solving for the only variable in the equation, they are solving for one variable of many. I will discuss later some ideas for getting over that hurdle; here, I started by examining the work Debbie showed:
Now, to solve this equation for r, the important thing is to make sure that at each step you are making a new equation that is still true--that is, an equivalent equation. We know that if we do the same thing to each side, e.g. adding the same thing, the new equation will be equivalent. But often students don't pay close attention to what they are doing, and they don't really make an equivalent equation.
Look closely at what you did:
M = C(1 + r)
-C -C
----------------
M - C = (1 + r)
What you say you are doing is subtracting C from both sides. But that's not really what you did. When you do the subtraction (without simplifying anything), you actually get
M - C = C(1 + r) - C
Debbie, like many students, had written “- C ” below the C, thinking that was the thing to do to get rid of the C, and then wrote below that what she expected the next line to be. You should always actually do what you say you are doing, to avoid this sort of mistake. By showing what the result of this subtraction really is, rather than (as we often do) just telling what she should have done, I am trying to show what it means to do what you say (that is, to really think about what you are doing), and also showing that “wrong” actions can be recovered from – everything you do doesn’t have to be the best thing to do, as long as it is a valid thing to do. So I continue from here:
Now you have to simplify, and you can't just cross off the C's! If we expand C(1 + r) using the distributive property, we get M - C = C + Cr - C and then combining like terms gives M - C = Cr That's not what you got! Why? Because you didn't actually subtract C from the right side, but just crossed something off, thinking it was the right thing to do. This is a very common mistake! You must always think of the subtraction as making a change to the equation and then simplifying, not just as canceling something.
Note that what she really did was to subtract C from the left, while dividing by C on the right. That does not lead to an equivalent equation, so her answer will be wrong.
We can continue, now--even though what you did is not the recommended method, which I will get to in a minute. Look at the equation we have now:
M - C = Cr
What is our goal? To get r by itself. We're almost there; all that's "wrong" with this is that r is multiplied by C, and we can get rid of that by dividing by C. So let's divide BOTH SIDES by C, again making sure that's really what we do:
M - C Cr
----- = --
C C
Now we can simplify the right side; multiplying r by C and then dividing by C undoes the multiplication and just leaves r:
M - C
----- = r
C
And that's our answer!
So her method led to a correct answer, though with more work than we might have needed.
A better method: simplify first
We can save work by thinking first about what will cause least trouble. Above, we saw that subtracting C required some hard work, because the C was not being added to something, but multiplied. Noticing what operation has to be undone is the key to efficient solving.
Now, there are two other methods that make the first step easier than what we ended up doing. One is to always simplify both sides of an equation before we start solving: M = C(1 + r) M = C + Cr (I distributed on the right side.)
(I sometimes describe this habit as “cleaning up the kitchen before you start to cook”, or something similar about a workbench. A neat working area, so to speak, makes it easier to see what you have to do, and leaves fewer places to make a mistake.)
Now we want to get r alone; on the right side it is FIRST being multiplied by C (remember the order of operations?) and THEN we're adding C to it. To get it by itself, we have to undo both operations; and we do that in reverse order. (For example, in the morning I put on my socks first, then my shoes; to undo that at night, I take off my shoes first, then my socks.) So we'll first undo the addition of C, by subtracting C from both sides:
M = C + Cr
- C - C
----- --------
M - C = Cr
What this really means is that we are subtracting C from both sides like this:
M - C = C + Cr - C = Cr
where I simplified the right side by combining the like terms C and -C.
Now we're right where we were in the first method, and just have to divide by C to finish.
When I fixed up Debbie’s work, we subtracted before dividing. In effect, we were taking off the sock before the shoe, which is harder but does end up with the same result (when done correctly). Luckily, algebraic socks don’t stretch out of shape!
Note that I used the same column format for my work as Debbie did, which is common in introducing algebra. Some authors avoid this notation, exactly because many students misuse it by not thinking about what it really means; they just write what I did afterward, on one line. I think the column format is excellent, as long as you think about it as meaning the same thing as the single line. And, of course, with experience one should not need to write everything out; students who continue rigidly writing everything out once they get past the introductory course are probably not thinking well.
Another good method: divide first
There's another way we could have done this, without simplifying first; we'd look at the equation as given and see that in M = C(1 + r) we are FIRST adding 1 to r, and THEN multiplying by C. We can undo that by FIRST dividing both sides by C, and THEN subtracting 1 from both sides. The answer we get would look different, but would mean the same thing. If you wish, you may try doing that; but the method I've shown is what is usually taught. I mention it because it is close to what you tried to do: you wanted to get rid of the C first; but because it is being MULTIPLIED rather than added, you have to DIVIDE by it rather than subtract, in order to eliminate it. What you really did was to subtract C from the left side and divide by C on the right, which didn't give an equivalent equation.
Here, by focusing on the operations being done, we see that (if we work with the form as given) we have to undo a multiplication first. Here is the work:
M = C(1 + r) --- -------- C C M/C = 1 + r M/C - 1 = r
Sometimes students now think they have the wrong answer, because it disagrees with the back of the book. But the same answer can be given in different, equivalent forms; we can show they are equivalent by rewriting one as the other. Students at this level often don’t know how to take this second form and turn it into the first; but we can turn the first into the second:
r = (M - C)/C = M/C - C/C = M/C - 1
What if you are stuck at the beginning?
Now, one thing that I often do with students struggling with formulas, that I didn’t do with Debbie because she wasn’t stumped about what to do (just wrong), is to suggest trying the problem first with only one variable, as a “dry run” for solving the formula. We can replace all the variables except the one we are solving for with simple numbers, and try solving in this more familiar setting:
M = C(1 + r)
becomes (arbitrarily replacing M with 5 and C with 2)
5 = 2(1 + r)
Most students know to first simplify, then subtract 2, and then divide by 2:
5 = 2 + 2r 3 = 2r 3/2 = r
Then I say, do the same things to the original form:
5 = s(1 + r) M = C(1 + r) 5 = 2 + 2r M = C + Cr 3 = 2r M - C = Cr 3/2 = r (M - C)/C = r
What we discover is that with variables, we can’t do the arithmetic, so all we can do to subtract is to write the subtraction (as an expression). This is new to the student, but I can point out that it really makes things easier — there’s no arithmetic to do!
We can also check our final result by plugging in the numbers we used for the dry run, M = 5 and C = 2: r = (5 – 2)/2 = 3/2. That brings more confidence.
One more thing I suggest, to avoid being overwhelmed by multiple variables, is to mark the one being solved for, to avoid losing track of the goal. I may circle or underline the r in the equation; and each time I move on to the next step, I first ask, What is our goal?
Solving formulas can be a Big New Thing, but mastering it solidifies your knowledge of the solving process.