As it has been in previous lessons, today's opener is a number riddle. What's unique about this one is that there are three unknown numbers to find. Where guess and check and graphing have been great options for solving previous problems, this one is designed to highlight the efficacy of algebraic substitution. Because I happen to be teaching this lesson on Pi Day, the numbers involved in this problem make a few very loose references to circles.
I give students a few moments to think about the problem. I want all students to be able to define the three variables and write equations, so as I circulate, I look at what they've got and in small group conversations I insist that they get at least that far. As for solving the problem, I recognize that some kids are going to need some practice with the algebraic steps, and that's what will happen over the course of the next few lessons.
After a few minutes I ask for everyone's attention and lead a discussion. I make sure that we've carefully and clearly defined the variables, and then the equations. As you can see in that image, we had some discussion about whether 4y = x or y = 4x would appropriately represent the second constraint, and posing this question is usually enough to give students the space to come to consensus.
As usual, I don't rush to a solution. I spend at most ten minutes, we get as far as we will, then I move on with the lesson. We'll come back to this. By not solving the problem (and for a math-loving teacher, it can be hard to do this), I build suspense and give kids a chance to grapple on their own. Without fail, some curious kids will approach at the end of each class to show me what they've got. This is a nice differentiation for the upper end of a class: give a problem that anyone can work on while the rest of the lesson is going on.
In terms of preparation, it will be great to be able look at a 3D graph of this system, but I don't force this. I'm ready for it, if it comes up, but I also know that it can be difficult for kids to really get what's going on if they're not motivated to understand it.
It's the end of the second week of this five-week Systems unit, and I have a few goals today. First of all, I want students to show me how much they can do with graphing as a method for solving systems problems. Second, I want to continue to illustrate the connections between, and relative strengths of, solving by graphing and solving by substitution. That second goal will be front and center on Monday, as we start a week of digging deeper into substitution, but it will continue to serve as a backdrop today.
Following the opener, I post the third slide of today's lesson notes. It looks like this:
I review the idea that SLT 5.2 is about solving problems in two variables by graphing, and that to do so requires a few background skills from earlier in the year. "First of all," I say, "you have to be able to graph linear equations. I know that all of you can do this, because I've seen you do it." At the very least, kids honed their linear graphing skills in the computer lab two days ago.
"Often, in order to have equations to be graphed," I continue, "you have to be able to take a problem and write equations to represent it. That's what SLT 1.5 says. Once you've mastered those two skills, all you really need to know is how to look for the intersection of two graphs, and to be able to interpret the solution. I'm going to show you what I mean right now."
A Break-Even Problem
To show everyone what I mean, I post the break-even problem on the fourth slide of today's lesson notes as an example. We work together to define the variables:
x = the number of t-shirts (produced or sold)
y = the amount money (spent or received)
and the equations
y = 1.25x + 72 (cost)
y = 5.75x (revenue)
I work quickly to lead the class through this part. I act like everyone is completely confident about this, to give everyone the feeling that they know this, that they've been practicing it, and that it's easy. If anyone is really stuck here, that will come out on the quiz at the end of class, and I'll proceed with those students accordingly.
Scaling the Axes
Once we have these equations, we can graph them. In contrast to the "Perimeter of a Rectangle" problem from yesterday's class, this problem will be easiest if we use different scales on each axis. On the x-axis, it will work to count by 1's (I'll justify this more or less depending on time constraints and whether or not kids press me on it, but there are more salient points to get to today). Counting by 1's will not suffice on the y-axis. We can see why immediately, by looking at the y-intercept of the cost function.
With the scales chosen, I give everyone a few moments to set up their axes. Then we graph. Unlike yesterday, this won't be as simple as plotting the y-intercept then counting boxes as indicated by the slope. But maybe there's a better way. To plot a line, I remind everyone, you don't need too many points. "In fact, what is the minimum number of points I need?" I ask. Almost everyone remembers that two is enough. I continue, "So what if we just figure out the cost of producing 0, 10, 20 and 30 t-shirts?" That will yield four simple enough points. Once we plot those points for each function, we can connect the dots for each, then find the intersection and answer the question.
Kids are in good shape at this point, and as I circulate, I tell them so. Then the really good part happens. Everyone starts to realize that it's hard to be perfectly confident about the solution they see at the intersection point. "I guess you'll have to make your best approximation of the solution here," I conclude.
So, What Can Substitution Tell Us?
Of course, this is where my second goal comes in to play. All we should really have to do is solve the equation
1.25x + 72 = 5.75x
and we'd be on our way. If anyone recognizes this, I give them props and tell them to keep quiet about that for now. That's where we're headed on Monday.
Here is the two-sided Graphing Quiz Template on which students will complete today's quiz. I like to use this structure because it allows for easy differentiation and re-use.
On the front of the quiz, students should graph and find the intersections of the four lines on the sixth slide of today's lesson notes. After projecting these on the screen for about ten minutes, I proceed to the problem on slide #7, which students should solve by graphing on the back of the quiz. This structure facilitates differentiation by class because it's easy to change the equations or the numbers in a problem on a slide. Offering another chance on a quiz like this is as easy as making new slides as well.
I've carefully chosen the four equations on the quiz with the same goal that I had during the break-even example in the previous section of this lesson. I want students to see that algebraic substitution is a useful complementary tool to graphing. As you can see below, three of the intersection points are lattice points. Those that are not consist of pairs of easily-managed decimals that work out nicely in an algebraic solution. I tell students that they have two options here: they can give approximate solutions when necessary, or they can use substitution to solve for the exact values of each point.
The second problem is a break-even problem that is just like the one we discussed earlier in class. I provide this assignment as a practice task as much as an assessment. I repeat the hint about scale as students move on to this problem, and tell students to show me as much as they can.
I want to ensure that there's a minimum of 20 minutes for kids to spend on this quiz. More is better. If there's less time than that, I'll save it for the next class. Yesterday's "Three Ways to Solve a Problem" was likely to extend a bit into today's lesson. If that was the case, it will make perfect sense to take three class periods to get through this and the previous lesson.