7.1
Solving
Systems of
Two Equations
Copyright © 2011 Pearson, Inc.
What you’ll learn about
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The Method of Substitution
Solving Systems Graphically
The Method of Elimination
Applications
… and why
Many applications in business and science can be
modeled using systems of equations.
Copyright © 2011 Pearson, Inc.
Slide 7.1 - 2
Solution of a System
A solution of a system of two equations in two
variables is an ordered pair of real numbers that
is a solution of each equation.
Copyright © 2011 Pearson, Inc.
Slide 7.1 - 3
Example Using the Substitution
Method
Solve the system using the substitution method.
2x - y = 10
6x + 4 y = 1
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Slide 7.1 - 4
Example Using the Substitution
Method
Solve the system using the substitution method.
2x - y = 10
6x + 4 y = 1
Solve the first equation for y.
2x - y = 10
y = 2x - 10
Substitute the expression for y into the second equation:
6x + 4(2x - 10) = 1
Copyright © 2011 Pearson, Inc.
Slide 7.1 - 5
Example Using the Substitution
Method
6x + 4(2x - 10) = 1
41
x=
14
y = 2x - 10
æ 41ö
= 2 ç ÷ - 10
è 14 ø
29
=7
æ 41 29 ö
The solution is the ordered pair ç ,- ÷ .
è 14 7 ø
Copyright © 2011 Pearson, Inc.
Slide 7.1 - 6
Example Solving a Nonlinear System
Algebraically
Solve the system algebraically.
y = x + 6x
2
y = 8x
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Slide 7.1 - 7
Example Solving a Nonlinear System
Algebraically
y = x 2 + 6x
y = 8x
Substitute the values of y from the first equation into
the second equation:
x 2 + 6x = 8x
x - 2x = 0
2
x = 0, x = 2.
If x = 0, then y = 0. If x = 2, then y = 16.
The system of equations has two solutions: (0,0) and (2,16).
Copyright © 2011 Pearson, Inc.
Slide 7.1 - 8
Example Using the Elimination
Method
Solve the system using the elimination method.
3x + 2 y = 12
4x - 3y = 33
Copyright © 2011 Pearson, Inc.
Slide 7.1 - 9
Example Using the Elimination
Method
Solve the system using the elimination method.
3x + 2 y = 12
4x - 3y = 33
Multiply the first equation by 3 and the second
equation by 2 to obtain:
9x + 6 y = 36
8x - 6 y = 66
Add the two equations to eliminate the variable y.
17x = 102
Copyright © 2011 Pearson, Inc.
so
x=6
Slide 7.1 - 10
Example Using the Elimination
Method
Solve the system using the elimination method.
3x + 2 y = 12
4x - 3y = 33
Substitue x = 6 into either of the two original equations:
3(6) + 2 y = 12
2 y = -6
y = -3
The solution of the original system is (6, - 3).
Copyright © 2011 Pearson, Inc.
Slide 7.1 - 11
Example Finding No Solution
Solve the system:
3x + 2 y = 5
-6x - 4 y = 10
Copyright © 2011 Pearson, Inc.
Slide 7.1 - 12
Example Finding No Solution
Solve the system:
3x + 2 y = 5
-6x - 4 y = 10
Multiply the first equation by 2.
6x + 4 y = 10
-6x - 4 y = 10
Add the equations:
0 = 20
The last equation is true for no values of x and y.
The equation has no solution.
Copyright © 2011 Pearson, Inc.
Slide 7.1 - 13
Example Finding Infinitely Many
Solutions
Solve the system.
3x + 6 y = -10
9x + 18y = -30
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Slide 7.1 - 14
Example Finding Infinitely Many
Solutions
Solve the system.
3x + 6 y = -10
9x + 18y = -30
Multiply the first equation by - 3.
-9x - 18y = 30
9x + 18y = -30
Add the two equations.
0=0
The last equation is true for all values of x and y.
The system has infinitely many solutions.
Copyright © 2011 Pearson, Inc.
Slide 7.1 - 15
Quick Review
1. Solve for y in terms of x. 2x + 3y = 6
Solve the equation algebraically.
2. x 3 = 9x
3. x 2 + 5x = 6
4. Write the equation of the line that contains the point
(1,1) and is perpendicular to the line 2x + 3y = 6.
5. Write an equation equivalent to x + y = 5 with
coefficient of x equal to - 2.
Copyright © 2011 Pearson, Inc.
Slide 7.1 - 16
Quick Review
2
1. Solve for y in terms of x. 2x + 3y = 6 y = - x + 2
3
Solve the equation algebraically.
2. x 3 = 9x
0,±3
3. x 2 + 5x = 6
- 6,1
4. Write the equation of the line that contains the point
(1,1) and is perpendicular to the line 2x + 3y = 6.
3
y - 1 = (x - 1)
2
5. Write an equation equivalent to x + y = 5 with
coefficient of x equal to - 2. - 2x - 2 y = -10
Copyright © 2011 Pearson, Inc.
Slide 7.1 - 17