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Chapter 4
Systems of
Linear Equations;
Matrices
Section
Section
Section
Section
Section
Section
Section
1
2
3
4
5
6
7
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–
–
–
–
–
–
Review: Sys of Linear Eg in Two Var
Sys of Linear Eq and Aug Matr
Gauss-Jordan Elimination
Matrices: Basic Operations
Inverse of a Square Matrix
Matrix Eq and Sys of Linear Eq
Leontief Input-Output Analysis
Introduction Chapter 4
 Systems of linear equations can be used to solve resource
allocation problems in business and economics.
 Such systems can involve many equations in many
variables.
• After reviewing methods for solving two linear equations
in two variables, we use matrices and matrix operations
to develop procedures that are suitable for solving linear
systems of any size.
• We also discuss the Wassily Leontief application of
matrices to economic planning of industrialized
countries.
Barnett/Ziegler/Byleen College Mathematics 12e
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Chapter 4
Systems of
Linear Equations;
Matrices
Section 1
Review: Systems of
Linear Equations in
Two Variables
Learning Objectives for Section 4.1
Review: Systems of Linear
Equations in Two Variables
 The student will be able to solve systems of linear equations in
two variables by graphing
 The student will be able to solve these systems using
substitution
 The student will be able to solve these systems using
elimination by addition
 The student will be able to solve applications of linear
systems.
Barnett/Ziegler/Byleen College Mathematics 12e
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Table of Content
 Systems of Linear Equations in Two
Variables
 Graphing
 Substitution
 Elimination by Addition
 Applications
Barnett/Ziegler/Byleen College Mathematics 12e
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Terms







solution
solution set
consistent / inconsistent system
dependent / independent system
substitution
elimination by addition
equilibrium price / quantity
Barnett/Ziegler/Byleen College Mathematics 12e
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Opening Example
A restaurant serves two types of fish dinners- small for $5.99
and large for $8.99. One day, there were 134 total orders of
fish, and the total receipts for these 134 orders was $1,024.66.
How many small fish dinners and how many large fish
dinners were ordered?
x fish dinner-small
y fish dinner-large
x+
y = 134
5.99x + 8.99y = 1,024.66
Barnett/Ziegler/Byleen College Mathematics 12e
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Opening Example
x fish dinner-small
y fish dinner-large
x+
y = 134
5.99x + 8.99y = 1,024.66
Now we have a system of two linear equations in two
variables. It is easy to find ordered pairs (x, y) that satisfies
one or the other of these equations. To solve this system we
have to find an ordered pair of real number that satisfies both
equation at the same time. In general, we have the following
definition:
Barnett/Ziegler/Byleen College Mathematics 12e
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Systems of Two Equations
in Two Variables
 We are given the linear system
ax + by = h
cx + dy = k
a, b, c, d, h, k real constants
A solution is an ordered pair (x0, y0) that will satisfy each
equation (make a true equation when substituted into that
equation). The solution set is the set of all ordered pairs
that satisfy both equations. In this section, we wish to find
the solution set of a system of linear equations. To solve a
system is to find its solution set.
Barnett/Ziegler/Byleen College Mathematics 12e
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Opening Example
We will consider three methods of solving such systems:
• Graphing
• Susbtitution and
• Elimination by addition
Barnett/Ziegler/Byleen College Mathematics 12e
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Solve by Graphing
 One method to find the solution of a system of linear
equations is to graph each equation on a coordinate plane
and to determine the point of intersection (if it exists). The
drawback of this method is that it is not very accurate in
most cases, but does give a general location of the point of
intersection. Lets take a look at an example:
 Solve the following system by graphing:
3x + 5y = –9
x+ 4y = –10
Barnett/Ziegler/Byleen College Mathematics 12e
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Solve by Graphing
Solution
3x + 5y = –9
x+ 4y = –10
First line (intercept method)
If x = 0, y= –9/5
If y = 0 , x = –3
Plot points and draw line
(2, –3)
Second line:
Intercepts are (0, –5/2) , ( –10,0)
From the graph we estimate that the point of intersection is (2,–3).
Check: 3(2) + 5(–3)= –9 and 2 + 4(–3)= –10. Both check.
Barnett/Ziegler/Byleen College Mathematics 12e
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Another Example
 Now, you try one:
 Solve the system by
graphing:
2x + 3 = y
x + 2y = –4
Barnett/Ziegler/Byleen College Mathematics 12e
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Another Example
Solution
 Now, you try one:
 Solve the system by
graphing:
2x + 3 = y
x + 2y = –4
First line (intercept method)
If x = 0, y= 3
If y = 0 , x = –3/2
Plot points and draw line
Barnett/Ziegler/Byleen College Mathematics 12e
Second line:
Intercepts are (0, –2) , ( –4,0)
From the graph we estimate
that the point of intersection is
(-2,–1).
14
Example 2
 Solve the following systems by graphing:
(A) x – 2y = 2
x+ y= 5
(B) x + 2y = -4
2x + 4y = 8
Barnett/Ziegler/Byleen College Mathematics 12e
(C) 2x + 4y = 8
x + 2y = 4
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Example 2 - Solution
(B) Lines are parallel
(each has slope -1/2) –
no solutions.
x + 2y = -4
2x + 4y = 8
(4,1)
(A) Interception at one point –
exactly one solution.
x – 2y = 2
x+ y= 5
(C) Lines coincide –
infinite number of
solutions.
2x + 4y = 8
x + 2y = 4
Barnett/Ziegler/Byleen College Mathematics 12e
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Basic Terms
 A consistent linear system is one that has one or more
solutions.
• If a consistent system has exactly one solution (unique
solution), it is said to be independent. An independent
system will occur when the two lines have different slopes.
• If a consistent system has more than one solution, it is said
to be dependent. A dependent system will occur when the
two lines have the same slope and the same y intercept. In
other words, the graphs of the lines will coincide. There
will be an infinite number of points of intersection. Two
systems of equations are equivalent if they have the same
solution set.
Barnett/Ziegler/Byleen College Mathematics 12e
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Basic Terms
 An inconsistent linear system is one that has no solutions.
This will occur when two lines have the same slope but
different y intercepts. In this case, the lines will be parallel
and will never intersect.
Referring to the three systems in the Example 2, the system in
part (A) is consistent and independent with the unique solution
x = 4, y = 1. The system in part (B) is inconsistent. And the
system in part (C) is consistent and dependent with an
infinitive number of solutions.
Barnett/Ziegler/Byleen College Mathematics 12e
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Example 2 - Solution
(B) Lines are parallel
(each has slope -1/2) –
no solutions.
x + 2y = -4
2x + 4y = 8
inconsistent
(4,1)
(A) Interception at one point –
exactly one solution.
x – 2y = 2
x+ y= 5
consistent and independent
(C) Lines coincide –
infinite number of
solutions.
2x + 4y = 8
x + 2y = 4
consistent and dependent
Barnett/Ziegler/Byleen College Mathematics 12e
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Basic Terms
 Example: Determine if the system is consistent,
independent, dependent or inconsistent:
2x – 5y = 6
–4x + 10y = –1
Barnett/Ziegler/Byleen College Mathematics 12e
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Solution of Example
 Solve each equation for y to obtain the slope intercept form of the
equation:
2𝑥 − 5𝑦 = 6
−4𝑥 + 10𝑦 = −1
−5𝑦 = −2𝑥 + 6
10𝑦 = 4𝑥 − 1
4
1
5𝑦 = 2𝑥 − 6
𝑦= 𝑥−
𝑦=
2
𝑥
5
−
6
5
𝑦=
10
2
𝑥
5
−
10
1
10
 Since each equation has the same slope but different y intercepts,
they will not intersect. This is an inconsistent system.
Barnett/Ziegler/Byleen College Mathematics 12e
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Basic Terms
 By graphing a system of two linear equations in two variables,
we gain useful information about the solution set of the
system. In general, any two lines in a coordinate plane must
intersect in exactly one point, be parallel, or coincide (have
identical graphs). The example 2 illustrate the only three
possible types of solutions for systems of two linear equations
in two variables. This is summarized as follows…
Barnett/Ziegler/Byleen College Mathematics 12e
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Possible Solutions to a Linear
System
 A linear system
ax + by = h
cx + dy = k
must have
A. Exactly one solution
or
B. No solution
Consistent and independent
Inconsistent
or
C. Infinitely many solutions
There are no other possibilities.
Barnett/Ziegler/Byleen College Mathematics 12e
Consistent and dependent
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Method of Substitution
 Although the method of graphing is intuitive, it is not very
accurate in most cases, unless done by calculator. There is
another method that is 100% accurate. It is called the
method of substitution. This method is an algebraic one.
It works well when the coefficients of x or y are either 1 or
–1. For example, let’s solve the previous system
2x + 3 = y
x + 2y = -4
using the method of substitution.
Barnett/Ziegler/Byleen College Mathematics 12e
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Method of Substitution
(continued)
The steps for this method are as follows:
1) Solve one of the equations for either x or y.
2) Substitute that result into the other equation to obtain an
equation in a single variable (either x or y).
3) Solve the equation for that variable.
4) Substitute this value into any convenient equation to
obtain the value of the remaining variable.
Barnett/Ziegler/Byleen College Mathematics 12e
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Example
(continued)
2x + 3 = y
x + 2y = –4
x + 2(2x + 3) = –4
x + 4x + 6 = –4
5x + 6 = –4
5x = –10
x = –2
Barnett/Ziegler/Byleen College Mathematics 12e
 Substitute y from first
equation into second
equation
 Solve the resulting equation
 After we find x = –2, then
from the first equation, we
have 2(–2) + 3 = y or y = –1.
Our solution is (–2, –1)
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Another Example
 Solve the system using
substitution:
3x – 2y = 7
y = 2x – 3
Barnett/Ziegler/Byleen College Mathematics 12e
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Another Example
Solution
 Solve the system using
substitution:
3x – 2y = 7
y = 2x – 3
 Solution:
3x  2 y  7
y  2x  3
3x  2(2 x  3)  7
3x  4 x  6  7
x  1
x  1
y  2  1  3  y  5
Barnett/Ziegler/Byleen College Mathematics 12e
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Elimination by Addition
 The method of substitution is not preferable if none of the
coefficients of x and y are 1 or –1. For example, substitution is
not the preferred method for the system below:
2x – 7y = 3
–5x + 3y = 7
 A better method is elimination by addition. The following
operations can be used to produce equivalent systems:
• Two equations can be interchanged.
• An equation can be multiplied by a non-zero constant.
• A constant multiple of one equation and is added to
another equation.
Barnett/Ziegler/Byleen College Mathematics 12e
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Elimination by Addition
Example
 For our system, we will seek to
eliminate the x variable. The
coefficients are 2 and –5. Our goal is
to obtain coefficients of x that are
additive inverses of each other.
 We can accomplish this by
multiplying the first equation by 5,
and the second equation by 2.
 Next, we can add the two equations
to eliminate the x-variable.
 Solve for y
 Substitute y value into original
equation and solve for x
 Write solution as an ordered pair
Barnett/Ziegler/Byleen College Mathematics 12e


5x  3y  7 
5(2x  7 y)  5(3) 

2(5x  3y){  2(7) 
10x  35y  15 

10x  6 y  14 
0x  29 y  29 
2x  7 y  3
y  1
2x  7(1) { 3 
2x  7  3 
2x  4
x  2
(2,1)
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Elimination by Addition
Another Example
Solve
2x – 5y = 6
–4x + 10y = –1
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Elimination by Addition
Another Example
2x – 5y = 6
–4x + 10y = –1
Solution:
1. Eliminate x by multiplying
equation 1 by 2 .
2. Add two equations
3. Upon adding the equations,
both variables are eliminated
producing the false equation
0 = 11
Solve
Barnett/Ziegler/Byleen College Mathematics 12e
2x  5 y  6


4 x  10 y  1
4 x  10 y  12 

4 x  10 y  1
0  11
4. Conclusion: If a false
equation arises, the system
is inconsistent and there is
no solution.
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Elimination by Addition
Additional Considerations
Let see what happens in the elimination process when a system
has either no solution or infinitely many solutions.
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Elimination by Addition
Additional Considerations
Consider the following system:
2x – 6y = -3
x + 3y = 2
Multiplying the second equation by -2 and adding, we obtain:
2x – 6y = -3
-2x – 6y = 2
0 = -7 not possible
We have obtained a contradiction. The initial assumption that the
system has solutions is false. The system has no solutions and
the graphs of the equations are parallel lines.
Barnett/Ziegler/Byleen College Mathematics 12e
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Elimination by Addition
Additional Considerations
Now consider the system:
x – ½y = 4
-2x + y = -8
Multiplying the top equation by 2 and adding, we obtain:
2x – y = 8
-2x + y = -8
0 = 0
This result implies that the equations are equivalent; that is, their
graphs coincide and the system is dependent.
Barnett/Ziegler/Byleen College Mathematics 12e
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Elimination by Addition
Additional Considerations
If we let x = k, where k is any real number, and solve either
equation for y, we obtain y = 2k – 8. So (k, 2k – 8) is a solution
to the system for any real number k.
The variable k is called a parameter and replacing k with a real
number produces a particular solution to the system.
For example, some particular solutions to the system are:
k = -1
k = -1
k = -1
k = -1
(-1, -10)
(2, -4)
(5, 2)
(9.4, 10.8)
Barnett/Ziegler/Byleen College Mathematics 12e
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Application
 A man walks at a rate of 3 miles per hour and jogs at a rate
of 5 miles per hour. He walks and jogs a total distance of
3.5 miles in 0.9 hours. How long does the man jog?
Barnett/Ziegler/Byleen College Mathematics 12e
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Application
 A man walks at a rate of 3 miles per hour and jogs at a rate
of 5 miles per hour. He walks and jogs a total distance of
3.5 miles in 0.9 hours. How long does the man jog?
 Solution: Let x represent the amount of time spent walking
and y represent the amount of time spent jogging. Since the
total time spent walking and jogging is 0.9 hours, we have
the equation x + y = 0.9.
We are given the total distance traveled as 3.5 miles. Since
Distance = Rate x Time, distance walking = 3x and
distance jogging = 5y. Then total distance is 3x + 5y= 3.5.
Barnett/Ziegler/Byleen College Mathematics 12e
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Application
(continued)
We can solve the system using  Solution:
substitution.
x  y  0.9
1. Solve the first equation for y
y  0.9  x
2. Substitute this expression into
3 x  5 y  3.5
the second equation.
3 x  5(0.9  x)  3.5
3. Solve second equation for x
3 x  4.5  5 x  3.5
4. Find the y value by substituting
this x value back into the first
2 x  1
equation.
x  0.5
5. Answer the question: Time spent
0.5  y  0.9
jogging is 0.4 hours.
y  0.4
Barnett/Ziegler/Byleen College Mathematics 12e
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Supply and Demand
The quantity of a product that people are willing to buy during
some period of time depends on its price. Generally, the higher
the price, the less the demand; the lower the price, the greater
the demand.
Similarly, the quantity of a product that a supplier is willing to
sell during some period of time also depends on the price.
Generally, a supplier will be willing to supply more of a
product at higher prices and less of a product at lower prices.
The simplest supply and demand model is a linear model
where the graphs of a demand equation and a supply equation
are straight lines.
Barnett/Ziegler/Byleen College Mathematics 12e
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Supply and Demand
(continued)
In supply and demand problems we are usually interested in
finding the price at which supply will equal demand. This is
called the equilibrium price, and the quantity sold at that
price is called the equilibrium quantity.
If we graph the the supply equation and the demand equation
on the same axis, the point where the two lines intersect is
called the equilibrium point. Its horizontal coordinate is the
value of the equilibrium quantity, and its vertical coordinate is
the value of the equilibrium price.
Barnett/Ziegler/Byleen College Mathematics 12e
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Supply and Demand
Example
Example: Suppose that the supply equation for long-life
light bulbs is given by
p = 1.04 q – 7.03,
and that the demand equation for the bulbs is
p = –0.81q + 7.5
where q is in thousands of cases. Find the equilibrium price
and quantity, and graph the two equations in the same
coordinate system.
Barnett/Ziegler/Byleen College Mathematics 12e
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Supply and Demand
(Example continued)
If we graph the two equations on a graphing calculator or
manually, or solve by substitution, and find the intersection
point, we see the graph below.
Demand Curve
1.04 q – 7.03 = –0.81q + 7.5
1.85 q = 14.53
q = 7.854
p = 1.04 q – 7.03
Supply Curve
p = 1.04 (7.854) – 7.03 = 1.14
Thus the equilibrium point is (7.854, 1.14), the
equilibrium price is $1.14 per bulb, and the equilibrium
quantity is 7,854 cases.
Barnett/Ziegler/Byleen College Mathematics 12e
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Now, Solve the Opening Example
 A restaurant serves two types of fish dinners- small for
$5.99 each and large for $8.99. One day, there were 134
total orders of fish, and the total receipts for these 134
orders was $1024.66. How many small dinners and how
many large dinners were ordered?
x fish dinner-small
y fish dinner-large
x+
y = 134
5.99x + 8.99y = 1,024.66
Barnett/Ziegler/Byleen College Mathematics 12e
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Solution
Multiply the first equation by -8.99 and add to the second equation:
(-8.99) (x + y) = 134(-8.99)
-8.99x - 8.99y = -1,204.66
5.99x + 8.99y = -1,024.66
-3x + 0 = -180.00
x = 60
x + y = 134
60 + y = 134
y = 134 – 60 = 74
Answer: 60 small orders and 74 large orders
Barnett/Ziegler/Byleen College Mathematics 12e
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Chapter 4
Systems of
Linear Equations;
Matrices
Section 1
Review: Systems of
Linear Equations in
Two Variables
END
Last Update: Abril 9/2013