Lesson#3: Solving Systems of Equations by the Substitution Method
Material Needed for Each Student:
8.5” x 11” paper
3” x 5” index cards
graphing calculator
Warm-up:
Imagine that you have a bag of 12 marbles: 3 blue, 4 red, and 5 green. You pick one
marble from the bag. What is the probability that it is red? What is the probability it is
red or blue? If you pick two marbles out of the bag, what is the probability they are both
green?
(Solution: Red marble: 3/12 = 1/4, Red or green: 7/12, Both green: 5/12 ∙ 4/11 = 20/132 =
5/33)
Chalkboard Example/Activity #1:
(1) x – 2y = 5
(2) 4x + 3y = 9
The first step is to write the first equation in terms of x: x = 2y + 5
On one side of the index card, the students will write x. On the other side, they will write
2y + 5 in parentheses. Then, the students write Equation 2 (4x + 3y = 9) in very large
letters across the 8.5” x 11”paper lengthwise. First, the students will place the index card
over the equation on the large paper so that the x on the index card is covering the x in 4x
+ 3y = 9 equation. Then, the students will flip over the index card. The large paper now
should read:
4(2y + 5) + 3y = 9
Now demonstrate the distribution property and solve for y.
8y + 20 + 3y = 9
11y + 20 = 9
11y = -11
y = -1
Distributive Property
Combine like terms
Divide by the coefficient
Now that the students have solved for y, the students will find x by substituting -1 for y in
both equations. (Side 1 of the classroom uses Equation 1 and Side 2 of the classroom
evaluates Equation 2.)
Side 1: x – 2y = 5
x – 2(-1) = 5
x+2=5
x=3
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Side 2: 4x + 3y = 9
4x + 3(-1) = 9
4x – 3 = 9
4x = 12
x=3
Since both sides of the room have solved x to be equal to 3, explain that after one
variable has been solved, the other variable can be found using either starting equation.
Since the students understand from the previous lesson that the solution to a system of
equations is a coordinate, explain that that the solution is to be written as (3,-1).
Next, show students how to check their answers by substituting (3, -1) for the variables in
each equation and verify that the answer is a true statement.
x – 2y = 5
3 – 2(-1) = 5
3+2=5
5=5
4x + 3y = 9
4(3) + 3(-1) = 9
12 – 3 = 9
9=9
Explain that each problem should be checked in this way to confirm the solution.
Chalkboard Example #2:
(1) 3x – 2y = x + 8
(2) y = x – 4
Repeat the index card method from the above example. The students now have:
3x – 2(x – 4) = x + 8
3x – 2x + 8 = x + 8
x+8=x+8
Substitute
Distributive Property
Combine like terms
Explain that this system of equations has infinite solutions, since the equation on eth left
is identical to the equation on the right. Then show that this means that the lines are
parallel. The answer is then: infinite solutions. Have the students solve each equation
for y and graph on the graphing calculator to confirm that the lines are identical.
3x – 2y = x + 8
-2y = -2x + 8
y = (-2x + 8)/-2
y=x-4
Chalkboard Example #3:
(1) 2x – y = 7
(2) 4x – 2y = 9
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First, explain to the students that in this example, the variable without any coefficients (y
in Equation 1) is negative and to take extra caution when solving for y.
2x – y = 7
-y = -2x + 7
y = 2x – 7
Using the index card method, students will write y on one side of the index card and
(2x – 7) on the other side. Then substitute the y in for the y on the large paper in
Equation 2 and flip the index card over.
4x – 2(2x – 7) = 9
4x – 4x + 14 = 9
14 = 9
Distributive Property
Combine like terms
At this point, explain that because the equations result in a false statement, the answer is
no solution. Now, have the students solve each equation for y and graph in their graphing
calculators.
2x – y = 7
-y = -2x + 7
y = 2x + 7
4x – 2y = 9
-2y = -4x + 9
y = (-4x + 9)/-2
When the students graph the equations, they will see that the lines are parallel and the
answer to this system is: no solution.
Chalkboard Example #4:
Loren’s marble jar contains plain marbles and colored marbles. If there are 32 more plain
marbles than colored marbles, and there are 180 marbles total, how many of each kind of
marble does she have?
First, explain to students that they will need to use the problem solving plan:
1) Read problem carefully. Decide what is given and what is unknown.
2) Assign variables to unknowns.
3) Write equations based on the given facts.
4) Solve the equation and find the numbers.
5) Check results and give the answer.
Now reread the problem and assign variables.
Let p = number of plain marbles
Let c = number of colored marbles
Write equations given the facts. Remind students that “more than” means addition and
“total” means to add both quantities together.
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p = 32 + c
p + c = 180
Now use the substitution method.
p + 32 = c
(32 + c) + c = 180
32 + 2c = 180
2c = 148
c = 74
Solve for p.
p = 32 + c
p = 32 + 74
p = 106
It seems that Loren has 74 colored marbles and 106 plain marbles. Check these amounts
with what is given in the problem and verify that it makes sense. Is 106 greater than 74
by 32? (Yes.) Is 74 + 106 = 180? (Yes.) The answer is 74 colored marbles and 106
plain marbles.
Guided Practice: Solve each system of equations using the substitution method.
During this activity, circulate and assist students as necessary.
1) 3x – 2y + 2 = 0
x + 3y = 14
2) 3x – 2y = 5
x + 2y = 15
3) x + 2y = 1
y=-½x+4
4) x – 4y = 8
4y – x + 8 = 0
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5) The length of a rectangle is 3 less than twice the width. The perimeter is 54. Find the
dimensions.
Recap: Review with students the steps to solving a system of equations by the
substitution method.
1) Write one of the equations in terms of one of the variables.
2) Substitute that into the other equation in place of the variable solved for in Step 1.
3) Solve the equation.
4) Substitute that value for the variable and solve for the missing variable in either
equation.
5) Write your solution as an ordered pair.
6) Check your answer!
Hand out assignment for individual practice.
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