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6.2 Using Substitution to
Solve Systems
Copyright © 2015, 2008 Pearson Education, Inc.
Section 6.2, Slide 1
Example: Making a Substitution for y
Solve the system
y  x3
2 x  3 y  14
Copyright © 2015, 2008 Pearson Education, Inc.
Equation (1)
Equation (2)
Section 6.2, Slide 2
Solution
From equation (1). We know that for any
solution of the system, the value of y is equal to
the value of x + 3. So, we substitute x + 3 for y
in equation (2):
2 x  3 y  14
2 x  3  x  3  14
Copyright © 2015, 2008 Pearson Education, Inc.
Equation (2)
Section 6.2, Slide 3
Solution
Note that, by making this substitution, we now
have an equation in one variable. Next, we solve
that equation for x:
2 x  3  x  3  14
2 x  3x  9  14
5x  9  14
5x  5
x 1
Copyright © 2015, 2008 Pearson Education, Inc.
Section 6.2, Slide 4
Solution
Thus the x-coordinate of the solution is 1. To
find the y-coordinate, we substitute 1 for x in
either of the original equations and solve for y:
y  x3
y  1 3
y4
Copyright © 2015, 2008 Pearson Education, Inc.
Equation (1)
Section 6.2, Slide 5
Solution
So, the solution is (1, 4). We can check that (1, 4)
satisfies both of the system’s equations:
Copyright © 2015, 2008 Pearson Education, Inc.
Section 6.2, Slide 6
Solution
Or, we can verify that (1, 4) is the solution by
graphing equations (1) and (2) and checking that
(1, 4) is the intersection point of the two lines. To
do so on a graphing calculator, we must first solve
equation (2) for y:
Copyright © 2015, 2008 Pearson Education, Inc.
Section 6.2, Slide 7
Solution
Copyright © 2015, 2008 Pearson Education, Inc.
Section 6.2, Slide 8
Using Substitution to Solve a Linear
System
To use substitution to solve a system of two linear
equations,
1. Isolate a variable on one side of either equation.
2. Substitute the expression for the variable found in
step 1 into the other equation.
3. Solve the equation in one variable found in step 2.
4. Substitute the solution found in step 3 into one of
the original equations, and solve for the other
variable.
Copyright © 2015, 2008 Pearson Education, Inc.
Section 6.2, Slide 9
Example: Isolating a Variable and Then
Using Substitution
Solve the system
3x  y  7
2x  5 y  1
Copyright © 2015, 2008 Pearson Education, Inc.
Equation (1)
Equation (2)
Section 6.2, Slide 10
Solution
We begin by solving for one of the variables in
one of the equations. We can avoid fractions by
choosing to solve equation (1) for y:
3x  y  7
Equation (1)
y  3x  7
Copyright © 2015, 2008 Pearson Education, Inc.
Section 6.2, Slide 11
Solution
Next, we substitute –3x – 7 for y in
equation (2) and solve for x:
2x  5 y  1
Equation (2)
2 x  5  3x  7   1
2 x  15x  35  1
17 x  35  1
17 x  34
x  2
Copyright © 2015, 2008 Pearson Education, Inc.
Section 6.2, Slide 12
Solution
Finally, we substitute –2 for x in the equation
y = –3x – 7 and solve for y:
y  3  2   7
y  67
y  1
The solution is (–2, –1). We could then verify
our work by checking that (–2, –1) satisfies
both of the original equations.
Copyright © 2015, 2008 Pearson Education, Inc.
Section 6.2, Slide 13
Example: Applying Substitution to an
Inconsistent System
Consider the linear system
y  3x  2
y  3x  4
Equation (1)
Equation (2)
The graphs of the equations are parallel lines
(why?), so the system is inconsistent and the
solution is the empty set. What happens when we
solve this system by substitution?
Copyright © 2015, 2008 Pearson Education, Inc.
Section 6.2, Slide 14
Solution
We substitute 3x + 2 for y in equation (2) and
solve for x:
y  3x  4
3x  2  3x  4
24
false
Equation (2)
We get the false statement 2 = 4.
Copyright © 2015, 2008 Pearson Education, Inc.
Section 6.2, Slide 15
Inconsistent System of Equations
If the result of applying substitution to a system of
equations is a false statement, then the system is
inconsistent; that is, the solution set is the empty
set.
Copyright © 2015, 2008 Pearson Education, Inc.
Section 6.2, Slide 16
Example: Applying Substitution to an
Dependent System
In an example in Section 6.1, we found that the
system
y  3x  5 Equation (1)
6 x  2 y  10 Equation (2)
is dependent and that the solution set is the infinite
set of solutions of the equation y = 3x – 5. What
happens when we solve this system by
substitution?
Copyright © 2015, 2008 Pearson Education, Inc.
Section 6.2, Slide 17
Solution
We substitute 3x – 5 for y in equation (2) and
solve for x:
6 x  2 y  10
6 x  2  3 x  5   10
6 x  6 x  10  10
10  10
true
Equation (2)
We get the true statement 10 = 10.
Copyright © 2015, 2008 Pearson Education, Inc.
Section 6.2, Slide 18
Dependent System of Equations
If the result of applying substitution to a system of
equations is a true statement that can be put into
the form a = a, then the system is dependent; that
is, the solution set is the set of ordered pairs
represented by every point on the (same) line.
Copyright © 2015, 2008 Pearson Education, Inc.
Section 6.2, Slide 19