Download Section 9.1

CHAPTER 9:
Systems of Equations
and Matrices
9.1
9.2
9.3
9.4
9.5
9.6
9.7
9.8
Systems of Equations in Two Variables
Systems of Equations in Three Variables
Matrices and Systems of Equations
Matrix Operations
Inverses of Matrices
Determinants and Cramer’s Rule
Systems of Inequalities and Linear Programming
Partial Fractions
Copyright © 2009 Pearson Education, Inc.
8.1
Systems of Equations in
Two Variables
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
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Solve a system of two linear equations in two variables by
graphing.
Solve a system of two linear equations in two variables using
the substitution and the elimination methods.
Use systems of two linear equations to solve applied
problems.
Copyright © 2009 Pearson Education, Inc.
Systems of Equations
A system of equations is composed of two or more
equations considered simultaneously.
Example: 5x  y = 5
4x  y = 3
This is a system of two linear equations in two
variables. The solution set of this system consists of
all ordered pairs that make both equations true. The
ordered pair (2, 5) is a solution of this system.
Copyright © 2009 Pearson Education, Inc.
Slide 9.1-4
Solving Systems of Equations
Graphically
When we graph a system of linear equations, each
point at which the graphs intersect is a solution of
both equations and therefore a solution of the
system of equations.
Copyright © 2009 Pearson Education, Inc.
Slide 9.1-5
Solving Systems of Equations
Graphically
Let’s solve the previous system graphically.
5x  y = 5
4x  y = 3
Solution:
We see that the graph
intersects at the single
point (2, 5), so this is the
solution of the system of
equations.
Copyright © 2009 Pearson Education, Inc.
Slide 9.1-6
Systems of Equations
If a system of equations has at least one solution, it
is consistent. If the system has no solutions, it is
inconsistent.
If a system of two linear equations in two variables
has an infinite number of solutions, the equations are
dependent. Otherwise, they are independent.
Copyright © 2009 Pearson Education, Inc.
Slide 9.1-7
Illustration of Graphs
Graphs of linear equations may be related to each
other in one of three ways.
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Slide 9.1-8
Substitution Method
The substitution method is a technique that gives
accurate results when solving systems of equations. It
is most often used when a variable is alone on one side
of an equation or when it is easy to solve for a variable.
One equation is used to express one variable in terms
of the other, then it is substituted in the other equation.
Copyright © 2009 Pearson Education, Inc.
Slide 9.1-9
Example
Use substitution to solve the system
5x  y = 5,
4x  y = 3.
Solution
Solve the first equation for y: y = 5x  5
Then we substitute 5x  5 for y in the second
equation to give an equation in one variable.
4x  (5x  5) = 3
4x  5x + 5 = 3
x=2
Copyright © 2009 Pearson Education, Inc.
Slide 9.1-10
Solution continued
Now we use back-substitution and substitute 2 for x
in either original equation.
4x  y = 3
4(2)  y = 3
8y=3
y=5
We find the solution to the system of equations to be
(2, 5), once again.
Copyright © 2009 Pearson Education, Inc.
Slide 9.1-11
Elimination Method
Using the elimination method, we eliminate one
variable by adding the two equations. If the
coefficients of a variable are opposites, that variable
can be eliminated by simply adding the original
equations. If the coefficients are not opposites, it is
necessary to multiply one or both equations by
suitable constants, before we add.
Copyright © 2009 Pearson Education, Inc.
Slide 9.1-12
Example
Solve the system using the elimination method.
6x + 2y = 4
10x + 7y =  8
Solution
If we multiply the first equation by 5 and the second equation
by 3, we will be able to eliminate the x variable.
30x + 10y = 20
Substituting:
6x + 2y = 4
30x  21y = 24
6x + 2(4) = 4
11y = 44
6x  8 = 4
y = 4
6x = 12
The solution is (2,  4).
x=2
Copyright © 2009 Pearson Education, Inc.
Slide 9.1-13
Another Example
Solve the system.
x  3y = 9 (1)
2x  6y = 3
(2)
Solution:
2x + 6y = 18
Mult. (1) by 2
2x  6y = 3
0 = 21
There are no values of x and y in which 0 = 21.
So this system has no solution. The graphs of the
equations are of parallel lines.
Copyright © 2009 Pearson Education, Inc.
Slide 9.1-14
Another Example
Solve the system.
9x + 6y = 48 (1)
3x + 2y = 16 (2)
Solution:
9x + 6y = 48
9x  6y = 48 Mult. (2) by 3
0=0
When we obtain the equation 0 = 0, we know the
equations are dependent. There are infinitely many
solutions. The graphs of the equations are identical.
Copyright © 2009 Pearson Education, Inc.
Slide 9.1-15
Application
Ethan and Ian are twins. They have decided to save
all of the money they earn, at their part-time jobs, to
buy a car to share at college. One week, Ethan
worked 8 hours and Ian worked 14 hours. Together
they saved $256. The next week, Ethan worked 12
hours and Ian worked 16 hours and they earned
$324. How much does each twin make per hour?
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Slide 9.1-16
Solution
Letting E represent Ethan and I represent Ian, the
following system can be obtained.
8E + 14I = 256
12E + 16I = 324
First week
Mult by 12
Mult by 8
96E + 168I = 3072
96E  128I = 2592
40I = 480
I = 12
Second week
Copyright © 2009 Pearson Education, Inc.
Slide 9.1-17
Solution
Solve for E.
8E + 14(12) = 256
8E = 88
E = 11
Ian makes $12 per hour while Ethan makes $11 per
hour.
Copyright © 2009 Pearson Education, Inc.
Slide 9.1-18