Copyright © 2007 Pearson Education, Inc. Slide 7-2
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 © 2007 Pearson Education, Inc. Slide 7-3
7.1 Systems of Equations
• A set of equations is called a system of equations.
• The solutions must satisfy each equation in the system.
• A linear equation in n unknowns has the form where the variables are of first-degree.
• If all equations in a system are linear, the system is a system of linear equations, or a linear system.
bxaxaxa nn 2211
Copyright © 2007 Pearson Education, Inc. Slide 7-4
• Three possible solutions to a linear system in two variables:
1. One solution: coordinates of a point,
2. No solutions: inconsistent case,
3. Infinitely many solutions: dependent case.
7.1 Linear System in Two Variables
Copyright © 2007 Pearson Education, Inc. Slide 7-5
7.1 Substitution Method
Example Solve the system.
Solution
4
31
1
55
11623
11)3(23
3
y
y
x
x
xx
xx
xy Solve (2) for y.
Substitute y = x + 3 in (1).
Solve for x.
Substitute x = 1 in y = x + 3.
Solution set: {(1, 4)}
3
1123
yx
yx (1)
(2)
Copyright © 2007 Pearson Education, Inc. Slide 7-6
7.1 Solving a Linear System in Two Variables Graphically
Example Solve the system graphically.
Solution Solve (1) and (2) for y.
3
1123
yx
yx (1)
(2)
Copyright © 2007 Pearson Education, Inc. Slide 7-7
7.1 Elimination Method
Example Solve the system.
Solution To eliminate x, multiply (1) by –2 and (2)
by 3 and add the resulting equations.
3696
286
yx
yx (3)
(4)
2
3417
y
y
1232
143
yx
yx (1)
(2)
Copyright © 2007 Pearson Education, Inc. Slide 7-8
7.1 Elimination Method
Substitute 2 for y in (1) or (2).
The solution set is {(3, 2)}.
• Check the solution set by substituting 3 in for x and 2 in for y in both of the original equations.
3
93
1)2(43
x
x
x
Copyright © 2007 Pearson Education, Inc. Slide 7-9
7.1 Solving an Inconsistent System
Example Solve the system.
Solution Eliminate x by multiplying (1) by 2 and
adding the result to (2).
Solution set is .
746
423
yx
yx (1)
(2)
746
846
yx
yx
150 Inconsistent System
Copyright © 2007 Pearson Education, Inc. Slide 7-10
7.1 Solving a System with Dependent Equations
Example Solve the system.
Solution Eliminate x by multiplying (1) by 2 and adding the result to (2).
Each equation is a solution of the other. Choose either equation and solve for x.
The solution set is e.g. y = –2:
42824
yxyx (1)
(2)
428
428
yx
yx
00
42
24
y
xyx
.,4
2
yy
)}2,1()}2,{( 422
Copyright © 2007 Pearson Education, Inc. Slide 7-11
7.1 Solving a Nonlinear System of Equations
Example Solve the system.
Solution Choose the simpler equation, (2), and solve for y since x is squared in (1).
Substitute for y into (1) .
43
523 2
yx
yx (1)
(2)
(3)3
4
43
xy
yx
34 x
Copyright © 2007 Pearson Education, Inc. Slide 7-12
7.1 Solving a Nonlinear System of Equations
Substitute these values for x into (3).
The solution set is
97
or 10)1)(79(
0729
15)4(29
53
423
2
2
2
xxxx
xx
xx
xx
1
314
or2743
34 9
7
yy
.1,1,, 2743
97
Copyright © 2007 Pearson Education, Inc. Slide 7-13
7.1 Solving a Nonlinear System Graphically
Example Solve the system.
Solution (1) yields Y1 = 2x; (2) yields Y2 = |x + 2|.
The solution set is {(2, 4), (–2.22, .22), (–1.69, .31)}.
(2)
(1)
02
2
yx
y x
Copyright © 2007 Pearson Education, Inc. Slide 7-14
7.1 Applications of Systems
• To solve problems using a system1. Determine the unknown quantities2. Let different variables represent those
quantities3. Write a system of equations – one for each
variable
Example In a recent year, the national average spent on two varsity athletes, one female and one male, was $6050 for Division I-A schools. However, average expenditures for a male athlete exceeded those for a female athlete by $3900. Determine how much was spent per varsity athlete for each gender.
Copyright © 2007 Pearson Education, Inc. Slide 7-15
7.1 Applications of Systems
Solution Let x = average expenditures per male y = average expenditures per female
100,1260502
yxyxAverage spent on
one male and one female
(2)
(1)
3900
12100
yx
yx
160002 x8000x
Average Expenditure per male: $8000, andper female: from (2) y = 8000 – 3900 = $4100.
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