(An archive question of the week)
I’m looking for past questions that led to deep discussions. This week, we have a case where a student realized he was doing algebra by rote, not thinking about what variables really mean. This realization was triggered by a step that many students stumble over, where parameters change their role. I have seen many calculus students struggle with this, so it deserves close examination.
Moving beyond routine
Here is the question, a long one from 2010:
Un-learning Unknowns Today, we started a unit on linear programming, but we first had a review of systems of equations, specifically for the upcoming SAT. We were looking at this problem: Each time Sue rents a bicycle, she pays a fixed base cost plus an hourly rate for the time the bicycle is rented. Last Saturday she paid $12.00 to rent a bicycle for 6 hours, and yesterday she paid $9.50 to rent a bicycle for 4 hours. Which of the following equations shows the total cost C, in dollars, for Sue to rent a bicycle for n hours? a) C = 1.25n b) C = 1.25n + 4.50 c) C = 2.00n + 2.50 d) C = 2.50n + 2.00 e) C = 4.50n + 1.25 First, I said ... 12.00 = m(6) + b ... where m is the slope of the line, or rate; and b is the fixed base price, or the y-intercept. Then, I said 9.50 = m(4) + b. Six and 4 are the number of hours. For the first equation, I set ... b = 12 - 6m ... and I substituted in this value into the second equation, so I got 9.50 = 4m + (12 - 6m). I then combined like terms and solved for m, arriving at m = 1.25. This made b = 4.50. Therefore, answer b) must be right.
The work is correct; Sam solved the system by substitution, and did it well. But then he stopped and thought about it, which is great:
But then I stepped back and thought for a moment. I have been doing systems of equations for so long that I have simply been going through the motions. As I looked over my work, I became more and more confused.
Please correct me if I am wrong, but I have always thought of systems of two-variable linear equations as lines to plot on a graph. The point where the lines intersect is the pair of coordinates that works for both graphs, and this is the solution to the system. For example, in y = 5x and y = -3x + 2, you are assuming that y and x are the same value, and this allows you to then substitute or eliminate. (It would be very helpful if you could comment or help me with this statement I just made, for I never fully understood how you can substitute y or x values if the equations are different.)
To solve this problem, I set up two equations. One was 12.00 = 6m + b and the other was 9.50 = 4m + b. My main confusion comes from the fact that a system of equations is the point where the two lines intersect. But how can you plot the lines here and see where they intersect? There is no y or x value, and you do not know m or b.
My second question came up when I wondered about plotting the lines. I visualized some lines, but I thought they were the same. What I am trying to say is that the two equations ...
12.00 = 6m + b
9.50 = 4m + b
... already have the same y-intercept and the same slope. They have different x-values, but the y-values correspond in a way that is proportional. If the two lines are the same, then how can you have an intersection point?
Essentially, the y-intercept and the slope of a line are the defining qualities of a line. (Right? Please help me with this.) If that is true, and we have two equations with the same slope and y-intercept, then how come they are not the same line? I mean ...
126 = y + 4x
96 = y + x
... are different because of the slope and y-intercept, right?
I am a slow learner, but I understand stuff when it is detailed, so please thoroughly explain the answer. THANK YOU SO MUCH because I am really confused right now.
Sam is overthinking; but the solution is not necessarily to back off and underthink! This is a wonderful question, because he wants to understand more deeply, and be able to think accurately.
Solving by substitution
I responded, starting with the fact that his work was correct:
Let me start by reassuring you that you did get the right answer. Now, on to your questions. Thinking about the meaning of what you're doing is a good thing!
First, we have to deal with the question,
“For example, in y = 5x and y = -3x + 2, you are assuming that y and x are the same value, and this allows you to then substitute or eliminate. (It would be very helpful if you could comment or help me with this statement I just made, for I never fully understood how you can substitute y or x values if the equations are different.)”
I answered, pointing out that his explanation was just right:
That's exactly it: you're assuming both equations are true (since you're assuming x and y are the solutions), so you can replace either side of one equation with the other side. That's what "equal" means -- they're the same value. So whichever you use, you get the same result.
The slope … of what?
His big problem starts when he says,
“But how can you plot the lines here and see where they intersect? There is no y or x value, and you do not know m or b.”
“What I am trying to say is that the two equations
12.00 = 6m + b
9.50 = 4m + b
already have the same y-intercept and the same slope. They have different x-values, but the y-values correspond in a way that is proportional.”
He is focusing on the names of the variables; here, m and b are a slope and a y-intercept, but not of these lines!
I continued,
We tend to use x and y so much that we forget that you can use ANY variables for the unknowns! In your case, the unknowns that you are looking for are not x and y, but m and b. So if you were to plot the lines, the axes would be labeled m and b, not x and y. Since m and b are now the unknowns, they are not the slope and intercept of the lines; they are the variables! The lines are ... 6m + b = 12.00 4m + b = 9.50 ... which you can think of as if they were 6x + y = 12.00 4x + y = 9.50 Each has its own slope (-6 and -4, respectively), and its own y-intercept (12 and 9.5, respectively) in terms of the variables m and b.
At this point in the problem, we are just solving a system of equations for variables m and b. We would not normally graph the equations, but would just solve in the abstract. Sam’s problem is in confusing this with one representation of the system, as a pair of lines. To see it that way, we have to set aside the meaning of the variables and just graph with one variable (I chose m) on the horizontal axis, and the other (b) on the vertical axis. Here is the result:
We can see that m = 1.25 and b = 4.5 at the intersection; so these values of m and b satisfy both equations. So the answer to the question is \(C = 1.25n + 4.50\), which is an entirely different line, where m and b have changed meanings. Here is that graph:
(Note, by the way, that what the problem is asking for can be thought of as finding the equation of the line through the points A(4, 9.50) and B(6, 12); this could have been done in several other ways, such as finding the slope as \(\displaystyle m = \frac{12 – 9.5)}{6 – 2} = \frac{2.5}{2} = 1.25\), and then using this in the point-slope form, \(y – y_1 = m(x – x_1)\). I put the two points on the graph to demonstrate that the solution is correct.)
This idea of taking the slope and the intercept from the original problem and turning them into the variables in a new problem (the system of equations) trips up a lot of people. Let algebra help you see abstractly: it doesn't matter what you call the variables in order to get this completely. Some textbooks, in fact, use a lot of unknowns other than x and y precisely so that students don't get too used to recognizing a kind of equation by the letters, but instead learn to understand them by the relationships they involve.
So Sam needs to avoid thinking of m and b as always meaning slope and y-intercept.
Once again, slowly
Sam wrote back:
Dr. Peterson,
Thank you for your answer, but I need a bit of clarification.
First, I am still a bit wary of the idea of changing the variables and equations. Do you think you could give me some examples? If so, that would be great.
Along with the problem I gave, when you say ...
6m + b = 12.00
4m + b = 9.50
... do you mean that the x-axis is the slope value and the y-axis is the y-intercept value? That would mean that a change in slope will cause a change in the y-intercept. I do not think this is correct, for how could you add the 6 times the slope if you already have a slope on the x-axis?
Also, with your example, how can you change the x-value that was given to you as the slope?
In addition, why are the lines ...
12.00 = 6m + b
9.50 = 4m + b
... not the same line? They both have the same slope and y-intercept.
I'm still confused, but I see where you might be going, and I thank you for your patience and help. Thank you for all you do at Dr. Math!!
I replied, offering a different example in the hope of breaking the connection to m and b:
As I said, this is a big hurdle for students, so I'm not at all surprised that you still have problems with it! Let's try a different approach. Here is a system of equations; can you solve it? 5a + 3b = 9 3a - 2b = 13 Of course you can. And when you do, you do not care at all whether I used a and b, or x and y, or some other pair of variables. The names don't matter; an equation is all about the relationship between the variables, regardless of their names or meanings. That is, it is abstract, and not tied to what the variables are. Let me know if you're okay so far, and I'll continue answering your specific questions after that.
Sam replied:
I think I am OK with that so far, but would you have to denote which variable would go on the x axis and which variable would go on the y axis?
Good question! I answered:
You COULD choose to graph the equations if you wish -- but the problem doesn't necessarily involve graphing. That is just one way to represent it. The problem itself is just about finding a pair of values that satisfies both equations. And if you did graph them, you could chose to put either variable on either axis. There is nothing inherent in the problem that makes one a better choice for the "independent variable" over the other. In other words, graphing is one TOOL you can use to interpret the problem -- namely, as the intersection of two lines. But the problem itself doesn't require that tool, either for solving it or for understanding it.
This is an essential point. A mathematical concept typically can be modeled in a number of different ways. We must not confuse the concept with one model. This is another aspect of the abstractness of mathematics.
I continued:
Now let's look back at your problem. You are trying to answer this question: Each time Sue rents a bicycle, she pays a fixed base cost plus an hourly rate for the time the bicycle is rented. Last Saturday she paid $12.00 to rent a bicycle for 6 hours, and yesterday she paid $9.50 to rent a bicycle for 4 hours. Which of the following equations shows the total cost C, in dollars, for Sue to rent a bicycle for n hours? That problem is not about a graph, but about rental costs. Your approach to solving it was to represent it as a problem about finding the equation of a line by looking for the slope and intercept. (There are other ways to model the problem, but you chose this because it is familiar. A standard method of problem solving is to model a new problem as one that is familiar, so you can use tools you already have.)
So Sam has shown that he can choose a representation for a problem. Good!
You also recognized that "a fixed base cost plus an hourly rate" means that the cost is a linear function. Suppose that function is C = mn + b or, using the variables you are more familiar with, y = mx + b Again, you think of it this way just because you want to turn the problem into one you know how to solve. All these variable names -- x, y, m, and b -- are things you chose to introduce, along with the idea of graphing. (By the way, the variables m and b here are parameters, which means they are considered fixed for the sake of the problem, namely graphing a line, whereas x and y are considered to actually vary. But really they are all just variables -- letters representing numbers you don't know. They just play different roles.)
Sam also recognized that the letters n and C played the roles commonly taken by x and y. The concept of a parameter may be the hardest here.
Parameters transformed
Now your problem becomes one of FINDING the parameters m and b, on the basis of two data points, namely that (x, y) = (6, 12) and (4, 9.5) are known to be points on the line y = mx + b. So you can write two equations by replacing x and y with those pairs:
12 = 6m + b
9.5 = 4m + b
This changes the models. Where a moment ago we had m and b as parameters describing a line (with x and y the variables), now they are unknown quantities you want to find! So you temporarily forget the original meaning of m and b, and just think of this as a new problem of solving a pair of equations.
It would look more typical, from your experience, if the equations were ...
6x + y = 12
4x + y = 9.5
... or ...
x + 6y = 12
x + 4y = 9.5
... depending on which of m and b you think of as "x." But it really makes no difference. The point is that m and b are now just unknown numbers we are trying to find. THEY ARE NOT A SLOPE AND AN INTERCEPT at this point; in particular, they are not the slope and intercept of these two lines! In fact, the slope of the line ...
6x + y = 12, or y = -6x + 12
... is -6. But that doesn't matter, because we don't even have to ask that question unless we choose to solve this system of equations by graphing. We just follow the usual methods to solve the equation and find what m and b are (namely 1.25, and 4.5). THEN we go back to the equation where we first used them, and put these values in their places:
C = mn + b
becomes
C = 1.25n + 4.5
This takes us all the way through Sam’s work.
Now let's answer the questions you asked last time:
Along with the problem I gave, when you say ...
6m + b = 12.00
4m + b = 9.50
... do you mean that the x-axis is the slope value
and the y-axis is the y-intercept value? That would
mean that a change in slope will cause a change in
the y-intercept. I do not think this is correct,
for how could you add the 6 times the slope if you
already have a slope on the x-axis?
If you chose to graph the equation 6m + b = 12, putting m on the horizontal axis and b of the vertical axis, then the graph would indeed show how the slope OF THE ORIGINAL EQUATION y = mx + b would force the y-intercept to change in order for the line to pass through (6, 12). That's because this equation represents the constraint that the line must pass through this point; if you imagine the line as a straight stick pinned to the plane at that point, and rotated that stick (changing the slope), it would change where it crossed the y-axis. For example, if m were 0, then the line would be y = 0x + 12 (so that y = 0*6 + 12 = 12 when x = 6), and so on.
But in solving the problem, you never stop to think about this level of meaning (and it doesn't matter) because at the point where you are working with the equation 6m + b = 12, m and b are nothing more than two numbers you are trying to find. The problem has been abstracted from its original context.
This is the power of algebra: we can turn any sort of problem into a problem that we can solve without having to think about the meaning of the quantities involved.
Here are graphs of several lines passing through (6, 12) at different slopes, showing how the y-intercept changes. If I increase the slope by 1, the intercept decreases by 6.
You also asked:
In addition, why are the lines ...
12.00 = 6m + b
9.50 = 4m + b
... not the same line? They both have the same slope and y-intercept.
As I've said this time, m and b are NOT the slope and y-intercept of either of these linear equations; they WERE the slope and y-intercept of a different line, y = mx + b or C = mn + b. But for the purposes of solving for m and b, they are just the two variables in this system of equations. We don't think of them as slope and intercept at this point, because what they were originally is irrelevant to the work we are doing. They are the slope and intercept of a line we are not currently paying any attention to.
Does that help at all?
It did. He answered:
Dear Dr. Peterson, I want to say thank you for all of your helpful responses. They really helped me understand the material and I appreciate the patience and depth of knowledge you have, along with all the other volunteers at Dr. Math!
That’s what we like to hear.