Solving Systems by Elimination
Lesson 18 of 20
Objective: SWBAT apply the idea that, given a system of two equations in two variables, replacing one equation by the sum of that equation and a multiple of the other produces a system with the same solution.
Opener: Must Have More Pie!
I want today's lesson to be fast-paced. Students will apply what they saw yesterday. Now that they have spent some time seeing that two equations can be combined by addition or subtraction, and that an optional intermediate step can be to multiply an equation by some factor, students will use that knowledge to solve some systems. I still like to call this the "elimination method" for solving systems, and that's the word I'll use here, but "addition method" or "combination method" can also work. For the sake of simplicity, I stick to the term "elimination" with my students.
In the previous problem, two students (I customize the names in the problem to each class) buy a whole bunch of cherry and pumpkin pies at a bake sale. The trick is that they both buy the same number of pumpkin pies, which gives rise to nice little shortcut.
As students get started, I ask them to compare this problem to the earlier pie problem. There's the obvious stuff: these two fellows are buying different kinds of pie, and they're spending more money. Then there's the observation that, this time, they're not buying the same number of either kind of pie. That makes this problem more difficult, which is important to acknowledge. Soon after that observation, lightning strikes! Someone comes up with the idea that if we could multiply "Joe's" purchase by three, then there would be 12 apple pies in each purchase, and then we could follow the same steps that we had previously.
I say that that's exactly the move we're going to make. That's what we saw yesterday, and it's what we're going to practice doing today. From here, I might show everyone what the algebra looks like, or we might just keep the solution informal. Whatever the format, we get the point where 35 strawberry pies cost $269.50, and the finish from there. The numbers might be bigger, and we might need a calculator to find 269.5/35, but the structure is just like that of the earlier problem.
One interesting side-note: kids are engaged enough that someone always asks whether we can multiply one whole equation by another. That's a fascinating question. I tell kids that it's a little harder, but to make sure that they take linear algebra when they get to college. I should also note that in the next unit, we'll be multiplying expressions by each other, and we'll see what happens there!
I especially like the approach to this lesson where we don't look at the algebraic solution to the opener, because that makes this next piece even more powerful. I put up the first of these two slides, and say, "How quickly can you solve this system?"
We've danced around the algebra for days, and now to see makes the elimination method really click for kids. They will often ask why didn't do this earlier, and I say that this method works best for linear equations in standard form, which is not what all linear equations look like when we first see them.
For the first problem, kids can move quickly, and if I know that they're already confident about this, I tell them to try to solve this system in their heads. Of course there are few sticky points here: we need to subtract the second equation from the first, and doing so will result in "subtracting a negative number," so we'll need to pay attention to that. Many students will also need to be reminded to find the value of x in the solution, not just y. If we didn't have to do so for y, it's often the solving for x part of this problem that requires us to write some things down.
The second of these two problems is like the second of the two pie problems. The steps to highlight here are choosing which equation needs to be multiplied, and what to multiply by. I ask, "What's easiest: turning a 2 into an 8 or turning a 3 into a 5?" The answer to that one is pretty obvious to kids. Ideally, someone will also wonder if we're allowed to multiply both questions, and in my experience, this happens about half the time. I say that we can, and that sometimes we're going to have to. It's a great habit to always try to find the most efficient way to solve a problem however, so whenever we have the option to use just one step, we should take it.
By now, everyone has a satisfying solution to the "Sweet Treats" problem on the front of the handout, but it can be useful to look at a solution by elimination.
Progress on the "Apples and Oranges" problem on the reverse is more mixed. If we need to, we'll spend a few minutes looking at the chart on the fifth slide of today's lesson notes. Here, we'll talk about some pattern that students have seen and why they're happening. I'm most interested in making sure to solve problem #5, which is reprinted on slide #6 of the lesson notes. Here, I'll make sure that kids see how to write a system of equations to solve this problem:
x + y = 68
0.55x + 0.75y = 40 OR 55x + 75y = 4000
At this point, I return to the strategy of leaving an unfinished problem. This is a challenging system to solve, and I want to give everyone room to try solving it on their own.
The same is true for problem #6, which really involves seeing a pattern in the table of possibilities.
I noted in the opener that I'm pushing for a fast-paced class today. I want students to see as much as they can of this new, but fairly intuitive, concept. Over the next few classes, they will then get to practice as much as necessary.
With that in mind, the final segment of today's class is on slides 7 through 13 of the lesson notes. I finally introduce the last Student Learning Target of Unit 5:
I can solve a system of equations by elimination.
I say that we've already been working on this, and that now I just want to make sure everyone knows what the learning target means. "When we combine two equations in such a way that one of the variables is eliminated, then we're using elimination to solve the system," I say. "I have three sets of four problems for you to try right now."
Take a look at slides 8 through 13. On each even-numbered slide are four systems, ready for solving by elimination. The odd-numbered slides have the solutions. I give students a certain amount of time to work on each set, usually 3 or 4 minutes on the first, and 5 or 6 on the third, depending on how much time is left in the class. If we don't get to the last set, that's fine. I want students to work hard under the time constraint, without feeling the pressure of a graded assessment. Because students can immediately check their work, and because they're racing the clock, this is productive time, and it frames the three-day practice session that will start tomorrow.