4.3
Solving Systems of Linear Equations by Substitution
Goal
Launch 4.3
• Develop and use the strategy for solving
If you did not use the more open alternate launch,
you can use the Getting Ready task to launch the
problem. By this point in the algebra strand, the
end of this worked example should not be
problematic for students. The difficult part is
likely to be the transition from step 1 to step 2.
In this transition, we need to change the way we
think about the linear equations in the system. In
general, we think of an equation of the form
y = mx + b as a dynamic relationship involving
many possible (x, y) pairs. When we have two
distinct, nonparallel linear relationships in a
system, there is one single point that solves the
system. When we solve a system, we are looking
for this one point; one (x, y) pair.
Help students understand that they are looking
for a single (x, y) pair. Then we can think of the
two equations in the system as putting constraints
on the x and y values. We are looking for an x and
y such that both equations are true.
When we substitute in step 2, we are
substituting equals for equals. There is some point
(x, y) such that y = 3x - 5, so we can substitute
3x - 5 for y. We are left with an equation only
in x, which students know how to solve.
Have students work in pairs.
systems of linear equations by substitution
One general technique for solving systems of
equations (not only linear systems) is to substitute
one expression for another. For solving linear
systems with two equations and two unknowns,
this substitution strategy is particularly helpful
when one of the equations already expresses one
variable in terms of the other or if it is easy to
transform one equation into that form.
Alternate Launch
You could launch the problem by challenging
students to find a third way to solve a system.
h
y 5 2x 1 5
4x 1 3y 5 25
Suggested Questions
• Here’s a system that is a mixture of the two
previous types. Can you think of yet a third
way one could solve systems, when they look
like this?
Explore 4.3
You might assign different systems in Question A
to different student pairs and ask them to prepare
a presentation of their results that shows the steps
in their reasoning.
For students working on the systems in
Question A, parts (4) and (5), if they are having
difficulty you may want to ask them why they
think that this problem is different than the one
they explored in Problem 4.2. Have them try to
graph the two equations in the system. Also
suggest they try another system from the
remaining parts of Question A.
Solving Systems of Linear Equations Symbolically
4
Investigation 4
I N V E S T I G AT I O N
You might try such a challenge to launch the
work on Problem 4.3, but the problem is also set
up to lead students more directly to useful
strategies through analysis of worked examples.
In the “analyzing-worked-examples” teaching
strategy, we ask the students to imagine that they
have come across someone else’s work on a
problem and to figure out whether the work is
correct and how the other person might have
been thinking about the problem and its solution.
85
Summarize 4.3
Have students present their solutions and
reasoning to the systems in task A.
Discuss Question C as a class.
• Would you use the substitution method or
some other method to solve the following
systems?
Suggested Question
• What does it mean to say that your point (x, y)
is a solution to the system?
• How do you know that your solution is
correct? How can you check?
• How is the substitution method different from
and similar to the method you used to solve
systems algebraically in Problems 4.1 and 4.2?
Discuss Question B and have the students who
worked on Question A, parts (4) and (5),
contribute to this discussion.
Have students give tips to solving a system
using the substitution method. Write these tips on
a piece of poster paper or on the board and have
students put the tips in their notebooks.
86
The Shapes of Algebra
h
4x 1 y 5 6
and
23x 1 y 5 1
h
2x 1 y 5 3
23x 2 7y 5 1
Answers will vary to this question, but a likely
answer is that the first system is easy to solve by
the method of Problem 4.2. This is because the
coefficient of y is 1 in both equations, so solving
for y is straightforward. However, both methods
are relatively easy to apply to this system. The
second system, however, results in inconvenient
fractions whether we solve for x or for y. In this
case, substitution keeps the numbers nicer and
may lead to fewer errors.
The choice of which method to use is a matter
of preference though some methods require fewer
steps. Observe students’ reasoning for why they
would choose to use a particular method.
At a Glance
4.3
Solving Systems of Linear Equations
by Substitution
PACING 1 day
Mathematical Goal
• Develop and use the strategy for solving systems of linear equations by
substitution.
Launch
Launch work on this problem with the Getting Ready task.
Materials
•
Help students to understand that they are looking for a single (x, y) pair.
Then we can think of the two equations in the system as putting constraints
on the x and y values. We are looking for an x and y such that both
equations are true.
Transparencies 4.3A,
4.3B
When we substitute in step 2, we are substituting equals for equals. Have
students work in pairs.
Explore
You might assign different systems in Question A to different student pairs
and ask them to prepare a presentation of their results that shows the steps
in their reasoning.
For students working on the systems in Question A, parts (4) and (5), if
they are having difficulty you may want to ask them why they think that
this problem is different than the one they explored in Problem 4.2. Have
them try to graph the two equations in the system. Also suggest they try
another system from the remaining parts of Question A.
Summarize
Have students present their solutions and reasoning to the systems in
Question A.
• What does it mean to say that your point (x, y) is a solution to the
Materials
•
•
Student notebooks
Poster paper
(optional)
system? How do you know that your solution is correct? How can you
check? How is the substitution method different from the method you
used to solve systems algebraically in Problems 4.1 and 4.2?
Discuss Question B and have the students who worked on Question A,
parts (4) and (5) contribute to this discussion.
Have students give tips to solving a system using the substitution
method. Write these tips on a piece of poster paper or on the board and
have students put the tips in their notebooks.
Discuss Question C as a class.
The choice of which method to use is a matter of preference though
some methods require fewer steps. Observe students’ reasoning for why
they would choose to use a particular method.
Investigation 4
Solving Systems of Linear Equations Symbolically
87
ACE Assignment Guide
for Problem 4.3
Core 15–20
Other 47–50, 63, and unassigned choices from
previous problems
Adapted For suggestions about adapting ACE
exercises, see the CMP Special Needs Handbook.
Connecting to Prior Units 47: Filling and Wrapping;
48, 49: Say It With Symbols; 50: Stretching and
Shrinking; 50: Kaleidoscopes, Hubcaps, and Mirrors
Answers to Problem 4.3
A. 1. Possible sequences:
Solve the first equation for y in terms of x.
x - 2(-2x - 1) = 12
5x + 2 = 12
x=2
y = -2(2) - 1
y = -5
Solve the second equation for x in terms
of y.
2(2y + 12) + y = -1
5y + 24 = -1
y = -5
This situation does not mean that any pair
(x, y) will satisfy the system, only that any
pair satisfying one of the equations will also
satisfy the other.
6. Solve either individual equation for x in
terms of y or vice versa. The solution is
x = 7.5 and y = 5.5.
B. 1. System (4) corresponds to two parallel lines
and system (5) corresponds to identical
lines.
h
2. In this system it is probably easier to solve
the first equation for y in terms of x, getting
y = -2x + 3. The solution to the system is
x = 2 and y = -1.
3. This system can be solved by substitution
by solving for x in terms of y or for y in
terms of x in the first equation or by solving
for x in terms of y in the second equation.
10
5
The solution is x = 2 3 and y = 3.
4. In this system it is probably easiest to solve
the first equation for y in terms of x. When
this relationship is substituted in the second
equation, the resulting equation
6x + 2(-3x + 4) = 7 yields
0x + 8 = 7. No values of x satisfy this
equation, so the system has an empty
solution set. This puzzling result can be
explained by graphing the two given
equations and discovering that the graphs
are parallel lines.
The Shapes of Algebra
This solution situation can be explained by
graphing the two lines and discovering that
the graphs are identical. Thus any ordered
pair (x, y) that satisfies one equation will
certainly satisfy the other.
2. In y = mx + b form, the systems are
x = 2(-5) + 12
x=2
88
5. In this system one could solve either
original equation for y in terms of x and
then substitute in the other. Trying the first
equation, one gets y = -1.5x + 5. The
substitution in the second equation yields
-6x - 4(-1.5x + 5) = -20. Proceeding
to solve this equation in x one arrives at
0x - 20 = -20 or simply -20 = -20.
Since this equation is true regardless of the
value of x, the equation has an infinite
number of solution pairs.
y 5 23x 1 4 and
y 5 23x 1 3.5
h
y 5 21.5x 1 5
y 5 21.5x 1 5
The first two lines are parallel and disjoint;
the second two are identical.
C. 1. See discussion of solution methods in the
Summarize section.
a. Q 7, 3 7 R
5
1
b. Q 17, 17 R
20 11
2. In system (a), it is easy to write each
equation in y = ax + b form. This makes
the equivalent form method useful.
In system (b), the arithmetic becomes more
tedious to use the equivalent form method.
However, the first equation is easy to write
in y = ax + b form and then to use
substitution.