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Copyright © 2005 Pearson Education, Inc.
SEVENTH EDITION and EXPANDED SEVENTH EDITION
Slide 7-1
Chapter 7
Systems of Linear Equations
and Inequalities
Copyright © 2005 Pearson Education, Inc.
7.1
Systems of Linear Equations
Copyright © 2005 Pearson Education, Inc.
System of Linear Equations
„
„
A solution to a system of equations is the
ordered pair or pairs that satisfy all equations in
the system.
A system of linear equations may have exactly
one solution, no solution, or infinitely many
solutions.
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Slide 7-4
Example: Is the Ordered Pair a
Solution?
„
„
Determine which of the ordered pairs is a
solution to the following system of linear
equations.
3x – y = 2
4x + y = 5
(2, 4)
(1, 1)
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Slide 7-5
Example: Is the Ordered Pair a
Solution? continued
Solution:
„
3x – y = 2
4x + y = 5
3(2) – 4 = 2
4(2) + 1 = 5
2 = 2 True
9 = 5 False
„
Since (2, 4) does not satisfy both equations, it is not a
solution to the system.
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Slide 7-6
Example: Is the Ordered Pair a
Solution? continued
„
„
3x – y = 2
3(1) – 1 = 2
2 = 2 True
4x + y = 5
4(1) + 1 = 5
5 = 5 False
Since (1,1) satisfies both equations, it is a
solution to the system.
Copyright © 2005 Pearson Education, Inc.
Slide 7-7
Procedure for Solving a System of
Equations by Graphing
„
„
„
Determine three ordered pairs that satisfy each
equation.
Plot the ordered pairs and sketch the graphs of both
equations on the same axes.
The coordinates of the point or points of intersection
of the graphs are the solution or solutions to the
system of equations.
Copyright © 2005 Pearson Education, Inc.
Slide 7-8
Graphing Linear Equations
When graphing linear equations, three solutions are
possible:
„
‰
The two lines may intersect at one point, producing a
system with one solution.
„
‰
A system that has one solution is called a consistent
system of equations.
The two lines may be parallel, producing a system with
no solutions.
„
A system with no solutions is called an inconsistent
system.
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Slide 7-9
Graphing Linear Equations continued
‰
The two equations may represent the same
line, producing a system with infinite number of
solutions.
„
A system with an infinite number of solutions is
called a dependent system.
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Slide 7-10
Example
„
„
Find the solution to the
following system of equations
graphically.
x – 3y = 9
2x – 6y = 3
Solution: Since the two lines
are parallel, they do not
intersect, therefore, the
system has no solution.
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Slide 7-11
7.2
Solving Systems of Equations
by the Substitution and
Addition Methods
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Procedure for Solving a System of
Equations Using the Substitution Method
„
„
„
„
Solve one of the equations for one of the variables. If
possible, solve for a variable with a coefficient of 1.
Substitute the expression found in step 1 into another
equation.
Solve the equation found in step 2 for the variable.
Substitute the value found in step 3 into the equation,
rewritten in step 1, and solve for the remaining
variable.
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Slide 7-13
Example: Substitution Method
„
Solve the following
system of equations by
substitution.
5x − y = 5
4x − y = 3
„
Solve the first equation
for y.
5x − y = 5
y = 5x − 5
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„
Then we substitute 5x − 5
for y in the other equation to
give an equation in one
variable.
4x − y = 3
4x − (5x − 5) = 3
Slide 7-14
Example: Substitution Method
continued
„
Now solve the equation
for x.
4x − 5x + 5 = 3
-x + 5 = 3
-x = -2
x=2
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„
„
Substitute x = 2 in the
equation solved for y and
determine the y value.
y = 5x − 5
y = 5(2) – 5
y = 10 – 5
y=5
Thus, the solution is the
ordered pair (2, 5).
Slide 7-15
Addition Method
„
„
If neither of the equations in a system of linear
equations has a variable with the coefficient of 1, it is
generally easier to solve the system by using the
addition (or elimination ) method.
To use this method, it is necessary to obtain two
equations whose sum will be a single equation
containing only one variable.
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Slide 7-16
Procedure for Solving a System of Equations
by the Addition Method
„
„
„
If necessary, rewrite the equations so that the
variables appear on one side of the equal sign and the
constant appears on the other side of the equal sign.
If necessary, multiply one or both equations by a
constant(s) so that when you add the equations, the
result will be an equation containing only one variable.
Add the equations to obtain a single equation in one
variable.
Copyright © 2005 Pearson Education, Inc.
Slide 7-17
Procedure for Solving a System of Equations
by the Addition Method continued
„
„
Solve the equation in step 3 for the variable.
Substitute the value found in step 4 into either
of the original equations and solve for the
other variable.
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Slide 7-18
Example: Multiplying Both Equations
Solve the system using the elimination method.
6x + 2y = 4
10x + 7y = − 8
„
Solution: In this system, we cannot eliminate a variable
by multiplying only one equation and then adding. To
eliminate the variable x, we can multiply the first
equation by 5 and the second equation by −3. We will
be able to eliminate the x variable.
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Slide 7-19
Continued, 10x + 7y = − 8
„
„
30x + 10y = 20
−30x − 21y = 24
− 11y = 44
y=−4
We can now find x by
substituting −4 for y in
either of the original
equations.
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6x + 2y = 4
„
Substituting:
6x + 2y = 4
6x + 2(−4) = 4
6x − 8 = 4
6x = 12
x=2
„
The solution is
(2, −4).
Slide 7-20
7.3
Matrices
Copyright © 2005 Pearson Education, Inc.
Matrix
„
A matrix is a rectangular array of elements.
‰
„
„
An array is a systematic arrangement of numbers or
symbols in rows and columns.
Matrices (the plural of matrix) may be used to display
information and to solve systems of linear equations.
The numbers in the rows and columns of a matrix are
called the elements of the matrix.
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Slide 7-22
Dimensions of a Matrix
„
„
The dimensions of a matrix may be indicated with the
notation r × s , where r is the number of rows and s is
the number of columns of a matrix.
A matrix that contains the same number of rows and
columns is called a square matrix.
‰
Example:
⎛ 2 5 7 ⎞ ⎛ 4 0 −2 ⎞
⎜
⎟⎜
⎟
3 x 3 square matrix ⎜ 1 2 6 ⎟ ⎜ 4 1 3 ⎟
⎜ 4 3 8⎟⎜6 8 9 ⎟
⎝
⎠⎝
⎠
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Slide 7-23
Addition and Subtraction of Matrices
„
„
Two matrices can only be added or subtracted if
they have the same dimensions.
The corresponding elements of the two matrices
are either added or subtracted.
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Slide 7-24
Adding Matrices
Example: Find A + B.
⎛ 5 7⎞
A=⎜
⎟,
⎝ −2 8 ⎠
Solution:
A+B
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⎛1 0⎞
B=⎜
⎟
⎝6 4⎠
⎛ 5 7⎞ ⎛ 1 0⎞
=⎜
⎟+⎜
⎟
2
8
6
4
−
⎝
⎠ ⎝
⎠
⎛ 5 +1 7 + 0⎞ ⎛6 7 ⎞
=⎜
⎟=⎜
⎟
⎝ −2 + 6 8 + 4 ⎠ ⎝ 4 12 ⎠
Slide 7-25
Multiplication of Matrices
„
„
„
A matrix may be multiplied by a real number, a scalar,
by multiplying each entry in the matrix by the real
number.
Multiplication of matrices is possible only when the
number of columns in the first matrix is the same as the
number of rows of the second matrix.
In general
⎛ a b ⎞ ⎛ e f ⎞ ⎛ ae + bg af + bh ⎞
A×B = ⎜
⎟⎜
⎟=⎜
⎟.
⎝ c d ⎠ ⎝ g h ⎠ ⎝ ce + dg cf + dh ⎠
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Slide 7-26
Example: Multiplying Matrices
Find A × B, given
⎛3 5⎞
⎛ −2 6 4 ⎞
A=⎜
⎟ and B = ⎜
⎟
4
1
0
3
7
⎝
⎠
⎝
⎠
„
Solution:
⎛ 3 5 ⎞ ⎛ −2 6 4 ⎞
A×B = ⎜
⎟⎜
⎟
4
1
0
3
7
⎝
⎠⎝
⎠
⎛ 3( −2) + 5(0) 3(6) + 5(3) 3(4) + 5(7) ⎞
=⎜
⎟
−
+
+
+
4(
2)
1(0)
4(6)
1(3)
4(4)
1(7)
⎝
⎠
⎛ −1 33 47 ⎞
=⎜
⎟
⎝ −7 27 23 ⎠
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Slide 7-27
Example: Identity Matrix in
Multiplication
Use the multiplicative identity matrix for a 2× 2
matrix and matrix A to show that
A × I = A.
„
⎡ 1 3⎤
A=⎢
⎥
⎣4 6⎦
⎛ 1 0⎞
Solution: The identity matrix is I = ⎜
⎟.
⎝ 0 1⎠
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Slide 7-28
Example: Identity Matrix in
Multiplication continued
⎡ 1 3 ⎤ ⎡ 1 0⎤
A ×I = ⎢
⎥
⎢
⎥
⎣4 6⎦ ⎣0 1⎦
⎡ 1(1) + 3(0) 1(0) + 3(1) ⎤
=⎢
⎥
⎣4(1) + 6(0) 4(0) + 6(1)⎦
⎡ 1 3⎤
=⎢
=A
⎥
⎣ 4 6⎦
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Slide 7-29
Multiplicative Identity Matrix
„
„
Square matrices have a multiplicative identity
matrix.
The following are the multiplicative identity for a
2 by 2 and a 3 by 3 matrix.
⎛ 1 0 0⎞
⎛ 1 0⎞
⎜
⎟
I =⎜
I = ⎜0 1 0⎟
⎟
⎝ 0 1⎠
⎜ 0 0 1⎟
⎝
⎠
‰
For any square matrix A, A × I = I × A = A.
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Slide 7-30
7.4
Solving Systems of Equations
by Using Matrices
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Augmented Matrix
„
The first step in solving a system of equations using
matrices is to represent the system of equations with an
augmented matrix.
‰
An augmented matrix consists of two smaller matrices,
one for the coefficients of the variables and one for the
constants.
Systems of equations
a 1 x + b 1y = c 1
a2x + b2y = c2
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Augmented Matrix
⎛ a1
⎜
⎝ a2
b1
b2
c1 ⎞
⎟
c2 ⎠
Slide 7-32
Row Transformations
„
„
To solve a system of equations by using matrices, we
use row transformations to obtain new matrices that
have the same solution as the original system.
We use row transformations to obtain an augmented
matrix whose numbers to the left of the vertical bar
are the same as the multiplicative identity matrix.
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Slide 7-33
Procedures for Row Transformations
„
„
„
Any two rows of a matrix may be interchanged.
All the numbers in any row may be multiplied by any
nonzero real number.
All the numbers in any row may be multiplied by any
nonzero real number, and these products may be
added to the corresponding numbers in any other row
of numbers.
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Slide 7-34
Example: Using Row Transformations
„
„
Solve the following system of
equations by using matrices.
x + 2y = 16
2x + y = 11
Solution: First we write the
augmented matrix.
⎛ 1 2 16 ⎞
⎜
⎟
⎝ 2 1 11 ⎠
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„
Our goal is to obtain a matrix
in the form:
⎛ 1 0 c1 ⎞
⎜
⎟
⎝ 0 1 c2 ⎠
„
It is generally easier to work in
columns. Since the element in
the top left position is already
1, we work to change the 2 in
the first column second row to
a zero.
Slide 7-35
Example: Using Row Transformations
continued
„
If we multiply the top row of
numbers by –2 and add these
products to the second row of
the second numbers, the
element in the first column,
second row will become 0.
‰ The top row of numbers
multiplied by –2 gives
1(−2), 2(−2) 16(−2)
„
Add these products to their
respective numbers in row 2.
⎛ 1 2 16 ⎞
⎜
⎟
⎝ 0 −3 −21⎠
„
The next step is to obtain a 1
in the second column, second
row. Multiplying all numbers in
the second row by − 1 .
3
⎛ 1 2 16 ⎞
⎜
⎟
0
1
7
⎝
⎠
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Slide 7-36
Example: Using Row Transformations
continued
„
The next step is to obtain a 0
in the second column first row.
Multiply the numbers in the
second row by −2 and add
these products to the
corresponding numbers in the
first row gives us 0.
„
With this desired augmented
matrix, we see that our
solution to the system is (2, 7).
„
This can be checked by
substituting these values into
the original equations.
⎛ 1 0 2⎞
⎜
⎟
0
1
7
⎝
⎠
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Slide 7-37
Inconsistent and Dependent Systems
„
An inconsistent system occurs when obtaining an
augmented matrix, one row of numbers on the left side
of the vertical line are all zeros but a zero does not
appear in the same row on the right side of the vertical
line.
‰
„
This indicates that the system has no solution.
If a matrix is obtained and a 0 appears across an entire
row, the system of equations is dependent.
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Slide 7-38
Triangularization Method
„
„
The triangularization method
can be used to solve a system
of two equations.
The ones and the zeros form a
triangle.
„
⎛ 1 2 16 ⎞
⎜
⎟
0
1
7
⎝
⎠
„
⎛1 a b⎞
⎜
⎟
0
1
c
⎝
⎠
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In the previous problem we
obtained the matrix
„
The matrix represents the
following equations.
x + 2y = 16
y=7
Substituting 7 for y in the
equation then solving for
x, x = 2.
Slide 7-39
7.5
Systems of Linear Inequalities
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Procedure for Solving a System of
Linear Inequalities
Select one of the inequalities.
„
‰
‰
Replace the inequality symbol with an equals sign and
draw the graph of the equation.
Draw the graph with a dashed line if the inequality is
< or > and with a solid line if the inequality is ≤ or ≥ .
Select a test point on one side of the line and
determine whether the point is a solution to the
inequality.
„
‰
If so, shade the area on the side of the line containing
the point. If the point is not a solution, shade the area
on the other side of the line.
Copyright © 2005 Pearson Education, Inc.
Slide 7-41
Procedure for Solving a System of
Linear Inequalities continued
„
„
Repeat steps 1 and 2 for the other inequality.
The intersection of the two shaded areas and
any solid line common to both inequalities
form the solution set to the system.
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Slide 7-42
Example: Solving a System of
Inequalities
Graph the following system of
inequalities and indicate the
solution set.
x + y >1
y ≤ 2x + 1
Solution: Graph both
inequalities on the same axis.
The first graph will have a
dashed line and the second
graph a solid line.
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Slide 7-43
Example: Another System of
Inequalities
Graph the following system of inequalities and indicate
the solution set.
x ≥ −3
y <1
Solution: Graph both on the same axes. The solution set
is the region of the graph that is shaded in both colors.
The point of intersection is not part of the solution
because it does not satisfy the inequality y<1.
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Slide 7-44
Example: Another System of
Inequalities continued
„
x ≥ −3
y <1
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Slide 7-45
7.6
Linear Programming
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Linear Programming
„
In a linear programming problem, there are restrictions
called constraints.
‰
‰
„
Each is represented as a linear inequality.
The list forms a system of linear inequalities.
When a system of inequalities is graphed, often a
feasible region is obtained.
‰
The points where two or more boundaries intersect are
called the vertices of the feasible region.
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Slide 7-47
Fundamental Principle of Linear
Programming
„
If the objective function, K = Ax + By, is
evaluated at each point in a feasible region, the
maximum and minimum values of the equation
occur at vertices of the region.
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Slide 7-48
Solving a Linear Programming
Problem
„
„
„
„
„
„
Determine all necessary constraints
Determine the objective function.
Graph the constraints and determine the feasible
region.
Determine the vertices of the feasible region.
Determine the value of the objective function at each
vertex.
The solution is determined by the vertex that yields
the maximum or minimum value of the objective
function.
Copyright © 2005 Pearson Education, Inc.
Slide 7-49
Example
„
Planer Carpentry makes rocking horses and rocking
airplanes. Each rocking horse requires 5 hours of
woodworking and 4 hours of finishing. Each airplane
requires 10 hours of woodworking and 3 hours of
finishing. Each month Planer Carpentry has 600 hours
of labor for woodworking and 140 hours for finishing.
The profit on each rocking horse is $40 and on each
airplane is $75. How many of each product should be
made in order to maximize profit?
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Slide 7-50
Example continued
„
Constraints
‰
‰
‰
‰
‰
‰
x = number of rocking horses
y = number of rocking airplanes
P = 40x + 75y (profit function)
5x + 10y ≤ 10 (woodworking hours)
4x + 3y ≤ 240 (finishing hours)
x ≥ 0 and y ≥ 0
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Slide 7-51
Example continued
Vertices
P = 40x + 75y
(0, 0)
P = 40(0) + 75(0) = 0
(60, 0)
P = 40(60) + 75(0) = 2400
(0, 60)
P = 40(0) + 75(60) = 4500
(24, 48)
P = 40(24) + 75(48) = 4560
The maximum profit would be from
making 24 rocking horses and 75
rocking airplanes.
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Slide 7-52