LE 11 Traditional Math: Substitution Method for Systems of Equations
Accelerated seventh grade and Math 8
Solving systems by substitution is generally a confusing topic for seventh graders as well as eighth graders. The approach I’ve used is to show how some systems lend themselves to substitution more than the elimination method. In this way, students see how substitution works as well as providing a simpler method for solving the problem.
Specifically, this pair of equations lends itself to substitution:
It is easily seen that four can be substituted for the y variable in the second equation. Similarly this pair also lends itself to substitution:
The first equation establishes that x + 4 can be used in place of y in the second equation. To solve by elimination would require transforming the top equation to y – x = 4, which in turn would require the second equation to be rewritten as y + x = 16.
On the other hand, systems of equations that lend themselves to elimination but which students are required to solve by substitution can be confusing. For example:
To solve by substitution, one possibility is for the first equation to be solved for x: x = 8 – 2y. The expression 8 – 2y is then substituted for x in the second equation. This appears straightforward to those who have had experience and are on the expert side of the novice/expert spectrum. But students encountering this for the first time, have difficulty making the substitution. Ultimately students will use the method of their choosing based on what appears to be the most straightforward route.
For this lesson, I focus on those problems that are suited for substitution. I include some that are more suitable for the elimination method as more challenging problems and offer extra credit for such problems on quizzes or tests, thus differentiating instruction.
Warm-Ups.
1. One number is 17 more than another. The sum of the numbers is 25. Find the numbers. (Let x be one number. Express the other number in terms of x and solve). Answer: x +x + 17 = 25; 2x + 17=25; 2x = 8; x = 4, x + 17 = 21
2. If y = 4, find the value of 5 + y.
3. Solve. 3x + 2(2x – 1)=12 Answer: This is a multi-step problem requiring distribution. Students may need help with this. 3x + 4x – 2 =16; 7x -2 = 12; 7x = 14; x = 2.
4. Solve the equations. y – x = 12; x + 2y = 36 Answer: Reverse terms for second equation so that it is 2y + x=48; 3y = 60; y = 20, x = 8
5. Solve the equations using elimination method. x + y = 21, x = y - 1 Answer: Rewrite second equation as x – y = -1; x = 10, y = 11.
Problem 1 is a word problem that has two unknowns and one variable. This is preparation for a later lesson on word problems using two variables for two unknowns. Problem 2 sets the stage for today’s lesson by using substitution to evaluate an expression. In the lesson students will take it a step further by substituting for a variable to solve an equation. Problem 3 is a multi-step problem requiring distribution which will be part of today’s lesson. Problem 4 may cause some confusion if homework problems did not include such a problem requiring terms to be reversed in one of the equations. Problem 5 will be used in this lesson as one that lends itself to solving by substitution.
The Wind-Up: Substitution Method: Monomials in Equations. In the Warm-Ups for the day’s lesson, Problem 2 has students plug in 4 for the expression y+5.
“You’ve done problems like that many times. It’s a plug-in, or substitution problem. Now let’s try something similar. You’re given that y = 6. Plug it in to x + y = 35 and solve.”
Students should be able to solve this with little prompting. If they need help, I ask “What does the equation look like after you’ve substituted 6 in for y? Can you solve it?” They should get 29.
“Let’s try something similar, but slightly different. Suppose I tell you y = 2x and to substitute this expression in x + y = 9. I want you to solve for x.” I have them work in their notebooks as I circulate around. I pick someone who has it right to put it on the board.
They should get x + 2x = 9, and getting x = 3.
Examples:
1. x = 3y. Substitute it in 5y – x = 8.” Answer: y = 4.
2. x = 5y; 2x + y = 22. This one is a bit different, so I go through it with them. “We are substituting 5y in for x, but x is being multiplied by 2. Fill in the blank:”
“If we fill in the blank we get:”
The Pitch: Substitution Method in Full. “In your Warm-Ups, Problem 5 had you solve by using elimination: x + y = 21, x = y – 1”
“You wound up with two equations that looked like: x + y = 21 and x – y = -1. Let’s do this a little differently. Instead of doing the extra step of making it solvable by elimination, I want you to substitute y – 1 in for x in the first equation.”
I write on the board the second equation but with a blank for x:
“Now replace the question mark in the box with y-1, write the new equation and solve it.”
They work in their notebooks as I walk around. They should have y – 1 + y = 21.
“Can you combine terms? What is the simplified version of the equation?”
Ultimately they should have 2y – 1 = 21, and y = 11.
“OK, now that you’ve solved for y, it’s just like you did before. Solve for x by substituting in either equation.” The answer is x = 10.
“This is the method of substitution. Sometimes equations lend themselves more to solving by substitution than they do elimination. Let’s do something more complex. Let’s say we have y = (x – 1); x + 2y = 13. This problem is similar to what we had before, except now we are substituting in an expression: y – 1. What are we substituting?”
I want to hear that we’re substituting (x-1) for y in the second equation.
“Let’s do that. Help me fill in the blank:
I walk around the room and pick someone who has filled in the blank correctly and have that person write the substituted equation on the board:
“We have a distribution going on; we’ve done these type of multi-step problems before; namely one of the Warm-Up problems. Let’s solve it.”
They should get x =5, y = 4.
Typical Area of Confusion: The main area of confusion is what gets substituted where. Students should look for the equation with the isolated variable as providing what gets substituted.
Examples: I work through the first two with them, and then have them work independently, providing guidance/answering questions as necessary. Most common error is students will forget to solve for the second variable.
1. x=2 – y; 2y + x = 9;Answer: 2y + 2 –y = 9; y = 7, x=-5
2. p=6 + 5q; 3p – 2q = 5;
Answer: 3(6 + 5q) – 2q = 5; 18 + 15q -2q = 5; 13q=-13; q = -1, p = 1
3. y = 2x -9; 5x + 2y = 27; Answer: 5x +2(2x-9)=27; 5x +4x – 18=27; 9x=45;
x=5, y=1
Solving for a Variable for Substitution. This last introduces the challenge type problems I mentioned in the introduction. I start out with: “If you saw this problem, how would you go about solving it?”
Students will likely say to add the two equations to cancel the y variable and solve.
“OK, but suppose I said to use substitution.”
There will be a combination of silence and then “It can’t be done.”
“Actually, it can be done. Look at the first equation. Can we solve for y?”
There will be general agreement.
“So now you have it in a form we’ve seen before: y = 5 -2x and 4x – y = 1. Can you substitute now?”
General agreement, but what follows is a dialogue reconstructed from various classes in which I’ve posed the above question.
“Why would we want to? It’s easy to solve with elimination.”
“Quite true, and if given a choice between substitution and elimination, which would you pick?”
Usually unanimous answer: “Elimination”
“Suppose I said on a quiz or test that you had to solve it by substitution.”
“Why would you do that?”
“Suppose I said that you could use any method you wanted, but I will give extra credit points if you use substitution.”
General silence and grudging agreement, and possibly “So we won’t have to do use substitution if we can use elimination?”
“If that’s the instruction on the test, yes.” With that understanding cemented in place, I proceed.
“Let’s do this one. x + y =3; 5x-3y = -1. “I want to get an equation that isolates either x or y. What should I do?” I’ll listen to suggestions; they will gravitate toward solving the first equation for either x or y. I’ll make it their choice and walk around the room to see their progress as they work it in their notebooks. They should get x = 1, y =2.
Homework. Homework should continue with the same type of problems. I’ll work one or two of the homework problems with the class as necessary to ensure they understand the procedures.