Transcript Document

6.3 Using Elimination to
Solve Systems
Copyright © 2015, 2008 Pearson Education, Inc.
Section 6.3, Slide 1
Adding Left Sides and Right Sides of
Two Equations
If a = b and c = d, then
a + c = b + d
In words, the sum of the left sides of two equations is
equal to the sum of the right sides.
Copyright © 2015, 2008 Pearson Education, Inc.
Section 6.3, Slide 2
Example: Solving a System by
Elimination
Solve the system
3x  2 y  9
3x  5 y  12
Copyright © 2015, 2008 Pearson Education, Inc.
Equation (1)
Equation (2)
Section 6.3, Slide 3
Solution
We begin by adding the left sides and adding the
right sides of the two equations:
3x  2 y  9
3x  5 y  12
0  7 y  21
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Equation (1)
Equation (2)
Section 6.3, Slide 4
Solution
By “eliminating” the variable x, we now have an
equation in one variable. Next, we solve that
equation for y:
0  7 y  21
7 y  21
y 3
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Section 6.3, Slide 5
Solution
Then, we substitute 3 for y in either of the
original equations and solve for x:
3x  2 y  9
3 x  2  3  9
3x  6  9
3x  3
x 1
Copyright © 2015, 2008 Pearson Education, Inc.
Equation (1)
Section 6.3, Slide 6
Solution
The solution is (1, 3). We check that (1, 3) satisfies
both equations (1) and (2):
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Section 6.3, Slide 7
Using Elimination to Solve a Linear
System
To use elimination to solve a system of two linear
equations,
1. Use the multiplication property of equality
(Section 4.3) to get the coefficients of one
variable to be equal in absolute value and
opposite in sign.
2. Add the left sides and add the right sides of the
equations to eliminate one of the variables.
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Section 6.3, Slide 8
Using Elimination to Solve a Linear
System
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 6.3, Slide 9
Example: Solving a System by
Elimination
Solve the system
5 x  4 y  13
9x  3 y  3
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Equation (1)
Equation (2)
Section 6.3, Slide 10
Solution
To eliminate the y terms, we multiply both sides
of equation (1) by 3 and multiply both sides of
equation (2) by 4, yielding the system
3(5 x  4 y  13)
4(9 x  3 y  3)
15x  12 y  39
36 x  12 y  12
Copyright © 2015, 2008 Pearson Education, Inc.
Section 6.3, Slide 11
Solution
The coefficients of the y terms are now equal in
absolute value and opposite in sign. Next, we add
the left sides and add the right sides of the
equations and solve for x:
15x  12 y  39
36 x  12 y  12
51x  0  51
51x  51
x 1
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Section 6.3, Slide 12
Solution
Substituting 1 for x in equation (1) gives
5 x  4 y  13
5 1  4 y  13
5  4 y  13
4 y  8
y  2
Equation (1)
The solution is (1, –2).
Copyright © 2015, 2008 Pearson Education, Inc.
Section 6.3, Slide 13
Solving Inconsistent Systems and
Dependent Systems by Elimination
If the result of applying elimination to a linear
system of two equations is
• a false statement, then the system is
inconsistent; that is, the solution set is the
empty set.
• 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.
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Section 6.3, Slide 14
Example: Solving a System by
Elimination
Solve the system
2x  7 y  5
6 x  21y  15
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Equation (1)
Equation (2)
Section 6.3, Slide 15
Solution
To eliminate the y terms, we multiply both sides
of equation (1) by –3, yielding the system
3(2 x  7 y  5)
6 x  21y  15
6 x  21y  15
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Section 6.3, Slide 16
Solution
Now that the coefficients of the x terms (and
those of the y terms) are equal in absolute value
and opposite in sign, we add the left sides and
add the right sides of the equations:
6 x  21y  15
6 x  21y  15
00  0
00
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Section 6.3, Slide 17
Solution
Since 0 = 0 is a true statement of the form a = a, we
conclude that the system is dependent and that the
solution set of the system is the set of ordered pairs
represented by the points on the line 2x – 7y = 5 and
the (same) line 6x – 21y = 15.
Copyright © 2015, 2008 Pearson Education, Inc.
Section 6.3, Slide 18
Solve a System of Two Linear
Equations
Any linear system of two equations can be
solved by graphing, substitution, or elimination.
All three methods will give the same result.
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Section 6.3, Slide 19