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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 7-2
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 
a x a x a x b n n     1 1 2 2 
first-degree. 
• If all equations in a system are linear, the system is a 
system of linear equations, or a linear system. 
Copyright © 2007 Pearson Education, Inc. Slide 7-3
7.1 Linear System in Two Variables 
• 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. 
Copyright © 2007 Pearson Education, Inc. Slide 7-4
7.1 Substitution Method 
Example Solve the system. 
Solution 
x y (1) 
3  2  
11 
 x  y 
 
3 
(2) 
y  x  
Solve (2) for y. 
x x 
3  2(  3)  
11 
x x 
3  2  6  
11 
x 
5 5 
1 
3 
 
 
  
1 3 
4 
 
x 
y 
y 
Substitute y = x + 3 in (1). 
Solve for x. 
Substitute x = 1 in y = x + 3. 
Solution set: {(1, 4)} 
Copyright © 2007 Pearson Education, Inc. Slide 7-5
7.1 Solving a Linear System in Two 
Variables Graphically 
Example Solve the system graphically. 
x y (1) 
3  2  
11 
 x  y 
 
3 
Solution Solve (1) and (2) for y. 
(2) 
Copyright © 2007 Pearson Education, Inc. Slide 7-6
7.1 Elimination Method 
Example Solve the system. 
x y (1) 
3  4  
1 
x y 
2  3  
12 
(2) 
Solution To eliminate x, multiply (1) by –2 and (2) 
by 3 and add the resulting equations. 
x y (3) 
    
6 8 2 
x y 
6  9  
36 
(4) 
y 
17 34 
2 
 
y 
 
Copyright © 2007 Pearson Education, Inc. Slide 7-7
7.1 Elimination Method 
Substitute 2 for y in (1) or (2). 
x 
3  4(2)  
1 
x 
3 9 
The solution set is {(3, 2)}. 
3 
 
 
x 
• Check the solution set by substituting 3 in for x 
and 2 in for y in both of the original equations. 
Copyright © 2007 Pearson Education, Inc. Slide 7-8
7.1 Solving an Inconsistent System 
Example Solve the system. 
x y (1) 
3  2  
4 
 6 x  4 y 
 
7 
(2) 
Solution Eliminate x by multiplying (1) by 2 and 
adding the result to (2). 
Solution set is . 
x y 
6  4  
8 
 x  y 
 
6 4 7 
0 15 Inconsistent System 
Copyright © 2007 Pearson Education, Inc. Slide 7-9
7.1 Solving a System with Dependent 
Equations 
Example Solve the system. 
x y (1) 
   
4 2 
x y 
8  2   
4 
(2) 
Solution Eliminate x by multiplying (1) by 2 and adding the 
result to (2). 
x y 
   
8 2 4 
x y 
8  2   
4 
0  0 
Each equation is a solution of the other. Choose either equation 
and solve for x. 
 
4 
x y x 
 4   2 
  
y 
2 
   
   
 
y 
  
The solution set is e.g. y = –2: 
2 
y 
, . 
4 
 
 
22     
{( , 2)} ( 1, 2)} 4 
Copyright © 2007 Pearson Education, Inc. Slide 7-10
7.1 Solving a Nonlinear System of 
Equations 
Example Solve the system. 
x y (1) 
3 2 
 2  
5 x y 
   
3 4 
(2) 
Solution Choose the simpler equation, (2), and 
solve for y since x is squared in (1). 
x y 
   
3 4 
 4 
 
 
3 
x 
y 
4x 
Substitute for y into (1) . 
(3) 
3 
Copyright © 2007 Pearson Education, Inc. Slide 7-11
7.1 Solving a Nonlinear System of 
Equations 
5 
  4 
 
3 
2 
x 
3  
2 
2 
 
  
 
 
 
x 
x x 
9  2(  4  )  
15 
2 
x x 
9  2  7  
0 
x x x x 
(9 7)( 1) 0 1 or 
       
Substitute these values for x into (3). 
 7 
   
  
4 9 
y  y 
3 
The solution set is 
7 
9 
  
1 
   
4 1 
3 
  or 
 
43 
27 
 7 ,  43 
,   1,  
1. 9 
27 
Copyright © 2007 Pearson Education, Inc. Slide 7-12
7.1 Solving a Nonlinear System 
Graphically 
Example Solve the system. 
(1) 
(2) 
y  
2 
x 
x y 
   
2 0 
Solution (1) yields Y1 = 2x; (2) yields Y2 = |x + 2|. 
The solution set is {(2, 4), (–2.22, .22), (–1.69, .31)}. 
Copyright © 2007 Pearson Education, Inc. Slide 7-13
7.1 Applications of Systems 
• To solve problems using a system 
1. Determine the unknown quantities 
2. Let different variables represent those 
quantities 
3. 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 7-14
7.1 Applications of Systems 
Solution Let x = average expenditures per male 
y = average expenditures per female 
x y 
 x y 
    
6050 12,100 
 
2 
Average spent on 
one male and one 
female 
(1) 
(2) 
12100 
3900 
x y 
  
x y 
  
2 x  16000 
x  8000 
Average Expenditure per male: $8000, and 
per female: from (2) y = 8000 – 3900 = $4100. 
Copyright © 2007 Pearson Education, Inc. Slide 7-15

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Systems of equations

  • 1.
  • 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 7-2
  • 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 a x a x a x b n n     1 1 2 2 first-degree. • If all equations in a system are linear, the system is a system of linear equations, or a linear system. Copyright © 2007 Pearson Education, Inc. Slide 7-3
  • 4. 7.1 Linear System in Two Variables • 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. Copyright © 2007 Pearson Education, Inc. Slide 7-4
  • 5. 7.1 Substitution Method Example Solve the system. Solution x y (1) 3  2  11  x  y  3 (2) y  x  Solve (2) for y. x x 3  2(  3)  11 x x 3  2  6  11 x 5 5 1 3     1 3 4  x y y Substitute y = x + 3 in (1). Solve for x. Substitute x = 1 in y = x + 3. Solution set: {(1, 4)} Copyright © 2007 Pearson Education, Inc. Slide 7-5
  • 6. 7.1 Solving a Linear System in Two Variables Graphically Example Solve the system graphically. x y (1) 3  2  11  x  y  3 Solution Solve (1) and (2) for y. (2) Copyright © 2007 Pearson Education, Inc. Slide 7-6
  • 7. 7.1 Elimination Method Example Solve the system. x y (1) 3  4  1 x y 2  3  12 (2) Solution To eliminate x, multiply (1) by –2 and (2) by 3 and add the resulting equations. x y (3)     6 8 2 x y 6  9  36 (4) y 17 34 2  y  Copyright © 2007 Pearson Education, Inc. Slide 7-7
  • 8. 7.1 Elimination Method Substitute 2 for y in (1) or (2). x 3  4(2)  1 x 3 9 The solution set is {(3, 2)}. 3   x • Check the solution set by substituting 3 in for x and 2 in for y in both of the original equations. Copyright © 2007 Pearson Education, Inc. Slide 7-8
  • 9. 7.1 Solving an Inconsistent System Example Solve the system. x y (1) 3  2  4  6 x  4 y  7 (2) Solution Eliminate x by multiplying (1) by 2 and adding the result to (2). Solution set is . x y 6  4  8  x  y  6 4 7 0 15 Inconsistent System Copyright © 2007 Pearson Education, Inc. Slide 7-9
  • 10. 7.1 Solving a System with Dependent Equations Example Solve the system. x y (1)    4 2 x y 8  2   4 (2) Solution Eliminate x by multiplying (1) by 2 and adding the result to (2). x y    8 2 4 x y 8  2   4 0  0 Each equation is a solution of the other. Choose either equation and solve for x.  4 x y x  4   2   y 2        y   The solution set is e.g. y = –2: 2 y , . 4   22     {( , 2)} ( 1, 2)} 4 Copyright © 2007 Pearson Education, Inc. Slide 7-10
  • 11. 7.1 Solving a Nonlinear System of Equations Example Solve the system. x y (1) 3 2  2  5 x y    3 4 (2) Solution Choose the simpler equation, (2), and solve for y since x is squared in (1). x y    3 4  4   3 x y 4x Substitute for y into (1) . (3) 3 Copyright © 2007 Pearson Education, Inc. Slide 7-11
  • 12. 7.1 Solving a Nonlinear System of Equations 5   4  3 2 x 3  2 2       x x x 9  2(  4  )  15 2 x x 9  2  7  0 x x x x (9 7)( 1) 0 1 or        Substitute these values for x into (3).  7      4 9 y  y 3 The solution set is 7 9   1    4 1 3   or  43 27  7 ,  43 ,   1,  1. 9 27 Copyright © 2007 Pearson Education, Inc. Slide 7-12
  • 13. 7.1 Solving a Nonlinear System Graphically Example Solve the system. (1) (2) y  2 x x y    2 0 Solution (1) yields Y1 = 2x; (2) yields Y2 = |x + 2|. The solution set is {(2, 4), (–2.22, .22), (–1.69, .31)}. Copyright © 2007 Pearson Education, Inc. Slide 7-13
  • 14. 7.1 Applications of Systems • To solve problems using a system 1. Determine the unknown quantities 2. Let different variables represent those quantities 3. 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 7-14
  • 15. 7.1 Applications of Systems Solution Let x = average expenditures per male y = average expenditures per female x y  x y     6050 12,100  2 Average spent on one male and one female (1) (2) 12100 3900 x y   x y   2 x  16000 x  8000 Average Expenditure per male: $8000, and per female: from (2) y = 8000 – 3900 = $4100. Copyright © 2007 Pearson Education, Inc. Slide 7-15