LE 9 Traditional Math: Solving Systems of Equations by Elimination
Accelerated Seventh Grade and Math 8
In going over the homework for the previous assignment, students undoubtedly ask about problems that have fractions as solutions, or have fractions in the equation. It becomes obvious that graphing is not the most efficient method for solving systems of equations. Also, even for problems for which the solutions are integers, the construction of the lines in the graph have to be done carefully.
In one class, I asked students what they thought of the method of graphing to solve equations. One boy said he liked it, and I asked why. “You can see what’s going on all at once,” he said. I found the answer insightful and remarked that indeed graphing does give you the big picture of what’s going on, even if it is not the best way to solve equations.
I move into solving systems of equations by the elimination method. Some textbooks (including the one I was using for seventh grade) introduce the substitution method first. Students find this method confusing, and I’ve found that teaching the elimination method first is a better introduction.
I split up the elimination method into two lessons. The first is for problems in which the elimination does not require transformation of any of the equations. That is, there are variables in both equations that can be cancelled by adding or subtracting the equations. For other equations in which there aren’t such variables (such as 3x + 2y = 10 and 5x – 3y = 4) one or both equations must be multiplied by an integer to allow for the elimination step to be carried out.
Warm-Ups.
1. A certain number plus twice that number is -24. Find the two numbers. Answer: Let x = a number; 2x = twice that number. Then x + 2x = -24; 3x =-24, x = -8, 2x = -16
2. Find the slope of the line that passes through (-3, 4) and ( -5, 4). Answer: m=0
3. -5 + 5 = ? Answer: 0
4. Without graphing, how many solutions does this system of equations have? y=2x-5, y-2x=6. Answer: None; second equation is y=2x + 6; the slopes are the same, so the lines are parallel.
5. Bob and Alicia’s weights average 57 pounds. If Alicia weighs 50 pounds, how much does Bob weigh? Answer: Total weight = 57 × 2 = 114 lbs. Bob’s weight: 114-50 =64 lbs
Problem 1 will be used later in this lesson to show how such problem can be solved with two variables. Problem 3 is there as a reminder that adding opposites equal zero, which is what happens in the elimination method.
Wind-Up: Adding the Same Amount to Each Side of an Equation. Students by now have learned that in solving equations we sometimes use the additive property of equality. I use this as a lead-in to the elimination method. After going through the homework, I ask the students if there were any things they did not like about the graphing method. Usually I hear that it takes time, it’s hard when you have fractions, and so on.
“Agreed. It isn’t the most efficient way of solving a system of equations. It does, however show what’s happening when we solve the equation. But now we want a more efficient method which is what we’ll be doing today.”
“Suppose I have the equations x + y = 5 and x – y = 2. I want to solve this but not by graphing. We all agree that we can add the same amount to both sides of an equation, do we not?”
Agreement ensues.
“So if I wanted to add 10 to both sides of the equation I could. It would look like this.”
I write the following on the board and students copy in their notebooks:
The Pitch: The Elimination Method. “I don’t really want to add 10 to both sides; I’m just making a point. We can add equal amounts to both sides of an equation. And since we can, there is something I’d like to add to both sides of the equation that will help me solve it. The other equation is x – y = 2. This means that the left side of the equation, x-y is the same number as 2, does it not? Isn’t that what the equal sign means?”
Usually there is nodding of heads but they are wondering where I’m going with this.
“Since I have an equal amount on both sides, I want to add this to the first equation:
“Can anyone tell me why I would want to do that?”
I usually get an answer along the lines that the y and –y will cancel.
“Exactly. If I add, the y’s disappear because we’re adding y to –y, and addition of opposites equal zero. In other words, they cancel. So let’s add the equations. What’s x + x? What’s 10 + 2?”
I now write on the board:
“Can we solve this?”
Students usually shout out the answer: x = 6. “Correct. So now that we know what x is, how can I use that to find out what y is?”
I hope for a student telling me I can plug it in to an equation. This may or may not happen. If it does not, I take the next card so to speak.
“Can I substitute 6 for x in the first equation, now that I know that’s what x equals? Let’s do it.” I write 6 + y = 10. “Can you solve for y?” They can, and they do; y = 4.”
“Let’s see if that works in the second equation: 6 -4 = 2. Yes, it works. That’s the answer.”
Once in a while, a student will say that the equation was so easy they were able to figure it out without going through all that; i.e., they used “guess and check”.
“Guess and check will work, yes, but it isn’t very efficient. Suppose we had these equations”:
“Would you want to use guess and check? I wouldn’t. Let’s try this one. What should we do? Tell me the steps.”
They tell me to add. “Why?” To cancel the y’s.
Adding we obtain 3x =240; x = 80. “Let’s plug this in to the top equation.”
Subtracting Equations. Let’s say we have this system:
Students will look at the x’s and say we should subtract.
“Yes, you can subtract, and that would be correct. But remember: Subtraction is the same thing as adding the opposite. And I find adding easier than subtracting, and I also find students make fewer mistakes when adding than subtracting. So I want to keep everything in addition mode. Let’s multiply one of the equations by -1.”
I show how we do this, picking the top equation, though I can pick either one. “We multiply each term by -1. So let’s do this.” I do it one step at a time, as what we get when we multiply by -1. We end up with –x – 4y = -27. Now we add the two equations. Do this in your notebooks.”
I circulate around to make sure students are doing this correctly, offering help when needed. The sum is -2y = -6. “And what do we get?”
Solving for y (and making sure they know to divide both sides by -2, not 2 which is a common mistake), we get y = 3. “Now substitute y in one of the original equations.” Again I walk around to see what they’re doing. They should get x = 15.
Examples. I work out the first few with the class so they are comfortable with the procedure.
1. x + y = 74, x - y = 16. Answer: x=45, y = 29
2. x + y = 122, x-y= 28; Answer: x = 75, y=47
In doing this problem students may, after solving for x plug it in to either equation to solve for y. I will show what happens if they plug it in to the second equation. They will get 75 – y = 28. Subtracting 75 from each side: -y = -47. I remind students that the coefficient of –y is -1, so both sides need to be divided by -1, leaving y= 47. This may be an area of confusion for students. In this case, they could plug x into the first equation and not run in to that situation. This may not always be an option, depending on what the equations are.
3. x + 6y = 10; x + 2y =2; I remind them that one of the equations should be multiplied by -1. I have them work it in their notebooks, and walk around offering guidance and then have someone work out the correct solution on the board.
Answer: x + 6y = 10; -x -2y = -2; 4y = 8; y = 2, x=-2
4. 3p – q = 10; 2p – q = 7. Answer: 3p-q = 10; -2p + q = -7; p = 3, q= -1
Homework. Problems are more of the same. As necessary I may work one or two of the homework problems as extensions of examples to be sure that students are comfortable with the procedure.