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1 Copyright © 2015, 2011, 2007 Pearson Education, Inc. Chapter 4-1 Systems of Equations and Inequalities Chapter 4

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Page 1: 1 Copyright © 2015, 2011, 2007 Pearson Education, Inc. Chapter 4-1 Systems of Equations and Inequalities Chapter 4

1 Copyright © 2015, 2011, 2007 Pearson Education, Inc. Chapter 4-1

Systems of Equations and Inequalities

Chapter 4

Page 2: 1 Copyright © 2015, 2011, 2007 Pearson Education, Inc. Chapter 4-1 Systems of Equations and Inequalities Chapter 4

2 Copyright © 2015, 2011, 2007 Pearson Education, Inc. Chapter 4-2

4.1 – Solving Systems of Linear Equations in Two Variables

4.2 – Solving Systems of Linear Equations in Three Variables

4.3 – Systems of Linear Equations: Applications and Problem Solving

4.4 – Solving Systems of Equations Using Matrices

4.5 – Solving Systems of Equations Using Determinants and Cramer’s Rule

4.6 – Solving Systems of Linear Inequalities

Chapter Sections

Page 3: 1 Copyright © 2015, 2011, 2007 Pearson Education, Inc. Chapter 4-1 Systems of Equations and Inequalities Chapter 4

3 Copyright © 2015, 2011, 2007 Pearson Education, Inc. Chapter 4-3

§ 4.1

Solving Systems of

Linear Equations in Two Variables

Page 4: 1 Copyright © 2015, 2011, 2007 Pearson Education, Inc. Chapter 4-1 Systems of Equations and Inequalities Chapter 4

4 Copyright © 2015, 2011, 2007 Pearson Education, Inc. Chapter 4-4

Definitions

When two or more linear equations are considered simultaneously, the equations are called a system of linear equations.

(1) y = x + 5(2) y = 2x + 4

A solution to a system of equations in two variables is an ordered pair that satisfies each equation in the system.The solution to the above system is (1,

6).

Page 5: 1 Copyright © 2015, 2011, 2007 Pearson Education, Inc. Chapter 4-1 Systems of Equations and Inequalities Chapter 4

5 Copyright © 2015, 2011, 2007 Pearson Education, Inc. Chapter 4-5

Solutions

Determine if (–4, 16) is a solution to the system of equations.

y = –4xy = –2x + 8

y = –4x16 = –4(–4)

16 = 16

y = –2x + 816 = –2(–4) + 8

16 = 8 + 816 = 16

Yes, it is a solution

Example:

Page 6: 1 Copyright © 2015, 2011, 2007 Pearson Education, Inc. Chapter 4-1 Systems of Equations and Inequalities Chapter 4

6 Copyright © 2015, 2011, 2007 Pearson Education, Inc. Chapter 4-6

Solutions

Determine if (–2, 3) is a solution to the system of equations.

x + 2y = 4y = 3x + 3

x + 2y = 4–2 + 2(3) = 4–2 + 6 = 4

4 = 4

y = 3x + 33 = 3(–2) + 3

3 = –6 + 33 = –3

But…

Example:

So it is NOT a solution

Page 7: 1 Copyright © 2015, 2011, 2007 Pearson Education, Inc. Chapter 4-1 Systems of Equations and Inequalities Chapter 4

7 Copyright © 2015, 2011, 2007 Pearson Education, Inc. Chapter 4-7

Types of Systems

The solution to a system of equations is the ordered pair (or pairs) common to all lines in the system when the system is graphed.

(–4, 16) is the solution to the system.

y = –4x

y = –2x + 8

Page 8: 1 Copyright © 2015, 2011, 2007 Pearson Education, Inc. Chapter 4-1 Systems of Equations and Inequalities Chapter 4

8 Copyright © 2015, 2011, 2007 Pearson Education, Inc. Chapter 4-8

Types of Systems

If the lines intersect in exactly one point, the system has exactly one solution and is called a consistent system of equations.

Page 9: 1 Copyright © 2015, 2011, 2007 Pearson Education, Inc. Chapter 4-1 Systems of Equations and Inequalities Chapter 4

9 Copyright © 2015, 2011, 2007 Pearson Education, Inc. Chapter 4-9

Types of Systems

If the lines are parallel and do not intersect, the system has no solution and is called an inconsistent system.

Page 10: 1 Copyright © 2015, 2011, 2007 Pearson Education, Inc. Chapter 4-1 Systems of Equations and Inequalities Chapter 4

10 Copyright © 2015, 2011, 2007 Pearson Education, Inc. Chapter 4-10

Types of Systems

If the two equations are actually the same and graph the same line, the system has an infinite number of solutions and is called a dependent system.

Page 11: 1 Copyright © 2015, 2011, 2007 Pearson Education, Inc. Chapter 4-1 Systems of Equations and Inequalities Chapter 4

11 Copyright © 2015, 2011, 2007 Pearson Education, Inc. Chapter 4-11

Solving GraphicallyTo Obtain a Solution to a System of Equations Graphically

Example: Solve the following system of equations graphically.

y = x + 2

y = -x + 4

1. Find the x- and y-intercepts.

Let x = 0; then y = 2. (0, 2)Let y = 0; then x = -2. (-2, 0)

Let x = 0; then y = 4. (0, 4)Let y = 0; then x =4. (4, 0)

2. Draw the graphs.

Graph each equation and determine the point or points of intersection.

Page 12: 1 Copyright © 2015, 2011, 2007 Pearson Education, Inc. Chapter 4-1 Systems of Equations and Inequalities Chapter 4

12 Copyright © 2015, 2011, 2007 Pearson Education, Inc. Chapter 4-12

Solving Graphically

Graph both equations on the same axes. The solutions is the point of intersection of the two lines, (1, 3).

Page 13: 1 Copyright © 2015, 2011, 2007 Pearson Education, Inc. Chapter 4-1 Systems of Equations and Inequalities Chapter 4

13 Copyright © 2015, 2011, 2007 Pearson Education, Inc. Chapter 4-13

Solve by Substitution

The substitution method for solving a system of equations can be used to find the solution to a system. The goal is to obtain one equation containing only one variable.

Page 14: 1 Copyright © 2015, 2011, 2007 Pearson Education, Inc. Chapter 4-1 Systems of Equations and Inequalities Chapter 4

14 Copyright © 2015, 2011, 2007 Pearson Education, Inc. Chapter 4-14

Solve by Substitution

1. Solve for a variable in either equation. If possible, solve for a variable with a numerical coefficient of 1 to avoid working with fractions.

2. Substitute the expression found for the variable in step 1 into the other equation. This will result in an equation containing only one variable.

3. Solve the equation obtained in step 2.4. Substitute the value found in step 3 into the equation from

step 1. Solve the equation to find the remaining variable. 5. Check your solution in all equations in the system.

To Solve a Linear System of Equations by Substitution

Page 15: 1 Copyright © 2015, 2011, 2007 Pearson Education, Inc. Chapter 4-1 Systems of Equations and Inequalities Chapter 4

15 Copyright © 2015, 2011, 2007 Pearson Education, Inc. Chapter 4-15

Solve by Substitution

Example: Solve the following system of equations by substitution.

y = 3x – 5y = -4x + 9

Page 16: 1 Copyright © 2015, 2011, 2007 Pearson Education, Inc. Chapter 4-1 Systems of Equations and Inequalities Chapter 4

16 Copyright © 2015, 2011, 2007 Pearson Education, Inc. Chapter 4-16

Solve by Substitution

2

147

957

9453

x

x

x

xx

Since both equations are already solved for y, we can substitute 3x – 5 for y in the second equation and then solve for the remaining variable, x.

Page 17: 1 Copyright © 2015, 2011, 2007 Pearson Education, Inc. Chapter 4-1 Systems of Equations and Inequalities Chapter 4

17 Copyright © 2015, 2011, 2007 Pearson Education, Inc. Chapter 4-17

Solve by Substitution

1

56

5)2(3

53

y

y

y

xy

Now find y by substituting x = 2 into the first equation.

Thus, we have x = 2 and y = 1, or the ordered pair (2, 1). A check will show that the solution to the system of equation is (2, 1).

Page 18: 1 Copyright © 2015, 2011, 2007 Pearson Education, Inc. Chapter 4-1 Systems of Equations and Inequalities Chapter 4

18 Copyright © 2015, 2011, 2007 Pearson Education, Inc. Chapter 4-18

Addition Method

The addition (or elimination) method for solving a system of equations can also be used to find the solution to a system. This method is generally the easiest one to use. Again, the goal is to obtain one equation containing only one variable.Example: Solve by using the addition method.

2x + 5y = 3 3x – 5y = 17

Page 19: 1 Copyright © 2015, 2011, 2007 Pearson Education, Inc. Chapter 4-1 Systems of Equations and Inequalities Chapter 4

19 Copyright © 2015, 2011, 2007 Pearson Education, Inc. Chapter 4-19

Addition Method 2x + 5y = 3 3x – 5y = 17

5x = 20 The y-variables are eliminated.

x = 4

Adding the equations yields: 2x + 5y = 3 3x – 5y = 17+

The x-variable can now be obtained using the same steps used for the substitution method.

Page 20: 1 Copyright © 2015, 2011, 2007 Pearson Education, Inc. Chapter 4-1 Systems of Equations and Inequalities Chapter 4

20 Copyright © 2015, 2011, 2007 Pearson Education, Inc. Chapter 4-20

Addition Method 2x + 5y = 3 3x – 5y = 17

Substitute x = 2 into either equation: 2x + 5y = 3 2(4) + 5y= 3

8 + 5y = 3 5y = -5

y = -1

The solution to this system is (4, -1).

Don’t forget to check your answer!

Page 21: 1 Copyright © 2015, 2011, 2007 Pearson Education, Inc. Chapter 4-1 Systems of Equations and Inequalities Chapter 4

21 Copyright © 2015, 2011, 2007 Pearson Education, Inc. Chapter 4-21

Addition Method Steps

1. If necessary, rewrite each equation in standard form, ax + by = c.

2. If necessary, multiply one or both equations by a constant(s) so that when the equations are added, the sum will contain only one variable.

3. Add the respective sides of the equations. This will result in a single equation containing only one variable.

4. Solve the equation obtained in step 3.

5. Substitute the value found in step 4 into either of the original equations. Solve that equation to find the value of the remaining variable.

6. Write the solution as an ordered pair.7. Check your solution in all equations in the system.

To Solve a System of Equations by the Addition Method

Page 22: 1 Copyright © 2015, 2011, 2007 Pearson Education, Inc. Chapter 4-1 Systems of Equations and Inequalities Chapter 4

22 Copyright © 2015, 2011, 2007 Pearson Education, Inc. Chapter 4-22

Addition Method

Example: Solve by using the addition method. 2x + y = 11 x + 3y = 18

y = 5

Since adding the equations at this point would not eliminate a variable, the first equation is multiplied by –2.

– 2(x + 3y = 18) + 2x + y =

11

– 2x - 6y = –36

Now the equations are added.

Page 23: 1 Copyright © 2015, 2011, 2007 Pearson Education, Inc. Chapter 4-1 Systems of Equations and Inequalities Chapter 4

23 Copyright © 2015, 2011, 2007 Pearson Education, Inc. Chapter 4-23

Addition Method

Example continued.

Now solve for x by substituting 5 for y in either of the original equations.

3

62

1152

112

x

x

x

yx

The solution is (3, 5).

Page 24: 1 Copyright © 2015, 2011, 2007 Pearson Education, Inc. Chapter 4-1 Systems of Equations and Inequalities Chapter 4

24 Copyright © 2015, 2011, 2007 Pearson Education, Inc. Chapter 4-24

Addition MethodExample: Solve by using the addition method.

x – 3y = 4 -2x + 6y = 1

0 = 9

Rewrite the equations so all the variables are on the left. Multiply equation 1 by 2.

+

Everything on the left cancels.

2x – 6y = 8-2x + 6y = 1

Since 0 =9 is a false statement, this system has no solution. The system is inconsistent and the graphs of these equations are parallel lines.

Set up the equations to add.– 2(x - 3y = 4)