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Slide 7.2-1
Chapter 7: Systems of Equations and
Inequalities; Matrices
7.1 Systems of Equations
7.2 Solution of Linear Systems in Three Variables
7.3 Solution of Linear Systems by Row Transformations
7.4 Matrix Properties and Operations
7.5 Determinants and Cramer’s Rule
7.6 Solution of Linear Systems by Matrix Inverses
7.7 Systems of Inequalities and Linear Programming
7.8 Partial Fractions
Copyright © 2011 Pearson Education, Inc.
Slide 7.2-2
7.2 Solution of Linear Systems in Three
Variables
• Solutions of systems with 3 variables with linear
equations of the form
Ax + By + Cz = D (a plane in 3-D space)
are called ordered triples (x, y, z).
• Possible solutions:
Copyright © 2011 Pearson Education, Inc.
Slide 7.2-3
7.2 Solving a System of Three Equations
in Three Variables
Solve the system.
3x  9 y  6 z  3
2x  y  z  2
(1)
(2)
x yz 2
(3)
Eliminate z by adding (2) and (3)
3x + 2y = 4 (4)
Eliminate z from another pair of equations, multiply (2) by 6
and add the result to (1).
12 x  6 y  6 z  12
3x  9 y  6 z  3
15 x  15 y  15
(5)
Eliminate x from equations 4 and 5.
Multiply (4) by -5 and add to (5).
15 x  10 y  10
15 x  15 y  15
5 y  5
y  1
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Slide 7.2-4
7.2 Solving a System of Three Equations
in Three Variables
continued
Using y = –1, find x from equation (4) by substitution.
3x + 2(–1) = 4
x=2
Substitute 2 for x and –1 for y in equation (3) to find z.
2 + (–1) + z = 2
z=1
The solution set is {(2, –1, 1)}.
Copyright © 2011 Pearson Education, Inc.
Slide 7.2-5
7.2 Solving a System of Two Equations
and Three Unknowns
Example
Solve the system.
x  2y 
z 
4
3x 
y  4z   9
(1)
(2)
Solution
Geometrically, the solution of two nonparallel planes is a line. Thus, there will be an infinite
number of ordered triples in the solution set.
 3R1  R 2  R 2
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x  2y  z 
4
 7 y  7 z   21
(1)
Slide 7.2-6
7.2 Solving a System of Two Equations
and Three Unknowns
 3R1  R 2  R 2
1
 R2  R2
7
x  2y  z 
4
 7 y  7 z   21
(1)
x  2y  z  4
y  z  3
(1)
This is as far as we can go with the echelon method.
Solve y + z = 3 to get y = 3 – z for any arbitrary
value for z. Now we express x in terms of z by
solving equation (1).
Copyright © 2011 Pearson Education, Inc.
Slide 7.2-7
7.2 Solving a System of Two Equations
and Three Unknowns
x  2y  z  4
x  2 y  z  4
x  2(3  z )  z  4
x  z2
Let y = 3 – z.
The solution set is written {(z – 2, 3 – z, z)}. For
example, if z = 1, then y = 3 – 1 = 2 and
x = 1 – 2 = –1, giving the solution set {(–1, 2, 1)}.
Verify that another solution is {(0, 1, 2)}.
Copyright © 2011 Pearson Education, Inc.
Slide 7.2-8
7.2 Application: Solving a System of Three
Equations to Satisfy Feed Requirements
An animal feed is made from three ingredients: corn, soybeans,
and cottonseed. One unit of each ingredient provides units of
protein, fat, and fiber, as shown in the table. How many units
of each ingredient should be used to make a feed that contains
22 units of protein, 28 units of fat, and 18 units of fiber?
Copyright © 2011 Pearson Education, Inc.
Slide 7.2-9
7.2 Application: Solving a System of Three
Equations to Satisfy Feed Requirements
Solution
Let x represent the number of units of corn,
y, the number of units of soybeans, and z, the number
of units of cottonseed. Using the table, we get the following
system
.25 x  .4 y  .2 z  22
Protein
.4 x  .2 y  .3z  28
Fat
.3x  .2 y  .1z  18
Fiber,
or,
25 x  40 y  20 z  2200
4 x  2 y  3z  280
3x  2 y 
z  180.
We can show that x = 40, y = 15, and z = 30.
Copyright © 2011 Pearson Education, Inc.
Slide 7.2-10
7.2 Curve Fitting Using A System
Find the equation of the parabola with vertical axis that
passes through the points (2, 4), (1, 1), and (2, 5).
Substitute each ordered pair into the equation ax2 + bx + c.
4 = 4a + 2b + c (1)
1=a–b+c
(2)
5 = 4a – 2b + c (3)
4  4a  2b  c
Eliminate c using equations (1) and (2). 1  a  b  c
3  3a  3b
Eliminate c using equations (2) and (3). 1 
a bc
5  4a  2b  c
4  3a  b
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Slide 7.2-11
7.2 Curve Fitting Using A System
3  3a  3b
continued
4  3a  b
Solve the system of equations in two 1  4b
variables by eliminating a.
1
 b
4
Find a by substituting for
b in equation (4).
1 a b
1
1 a 
4
5
a
4
Find c by substituting
for a and b in equation (2)
1 a bc
Find c by substituting a and b in equation (2).
The equation of the parabola is: y 
Copyright © 2011 Pearson Education, Inc.
5  1
 c
4  4
6
1 c
4
1
 c
2
1
5 2 1
1
x  x .
4
4
2
Slide 7.2-12