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Chapter 4

Systems of Linear

Equations and Inequalities

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Systems of Equations and Inequalities

4.1 Solving Systems of Linear Equations in Two Variables

4.2 Solving Systems of Linear Equations in Three Variables

4.3 Solving Applications Using Systems of Equations

CHAPTER

44

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Solving Systems of Linear Equations in Three Variables

1. Determine if an ordered triple is a solution for a system of equations.

2. Understand the types of solution sets for systems of three equations.

3. Solve a system of three linear equations using the elimination method.

4.24.2

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ExampleDetermine whether (2, –1, 3) is a solution of the system 4,

2 2 3,

4 2 3.

x y z

x y z

x y z

Solution In all three equations, replace x with 2, y with –1, and z with 3.

x + y + z = 4

2 + (–1) + 3 = 4

4 = 4

TRUE

3 = 3

TRUE

2x – 2y – z = 3

2(2) – 2(–1) – 3 = 3

– 4x + y + 2z = –3

– 4(2) + (–1) + 2(3) = –3

–3 = –3

TRUE

Because (2, 1, 3) satisfies all three equations in the system, it is a solution for the system.

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Types of Solution Sets A Single Solution: If the planes intersect at a single point, that ordered triple is the solution to the system.

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Types of Solution Sets Infinite Number of Solutions: If the three planes intersect along a line, the system has an infinite number of solutions, which are the coordinates of any point along that line.

Infinite Number of Solutions:

If all three graphs are the same plane, the system has an infinite number of solutions, which are the coordinates of any point in the plane.

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Types of Solution Sets No Solution: If all of the planes are parallel, the system has no solution.

No Solution: Pairs of planes also can intersect, as shown. However, because all three planes do not have a common intersection, the system has no solution.

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Example

Solution

Solve the system using elimination.

6,

2 2,

3 8.

x y z

x y z

x y z

We select any two of the three equations and work to get one equation in two variables. Let’s add equations (1) and (2):

(1)

(2)

(3)

6

2 2

x y z

x y z

(1)

(2)

(4)2x + 3y = 8Adding to eliminate z

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Next, we select a different pair of equations and eliminate the same variable. Let’s use (2) and (3) to again eliminate z.

2 2

3 8

x y z

x y z

(5) x – y + 3z = 8

Multiplying equation (2) by 3

4x + 5y = 14.

3x + 6y – 3z = 6

Now we solve the resulting system of equations (4) and (5). That will give us two of the numbers in the solution of the original system,

4x + 5y = 14

2x + 3y = 8(5)

(4)

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We multiply both sides of equation (4) by –2 and then add to equation (5):

Substituting into either equation (4) or (5) we find that x = 1.

4x + 5y = 14–4x – 6y = –16,

–y = –2 y = 2

Now we have x = 1 and y = 2. To find the value for z, we use any of the three original equations and substitute to find the third number z.

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Let’s use equation (1) and substitute our two numbers in it:

We have obtained the ordered triple (1, 2, 3). It should check in all three equations.

x + y + z = 6

1 + 2 + z = 6z = 3.

The solution is (1, 2, 3).

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Solving Systems of Three Linear Equations Using Elimination

1. Write each equation in the form Ax + By+ Cz = D.

2. Eliminate one variable from one pair of equations using the elimination method.

3. If necessary, eliminate the same variable from another pair of equations.

4. Steps 2 and 3 result in two equations with the same two variables. Solve these equations using the elimination method.

5. To find the third variable, substitute the values of the variables found in step 4 into any of the three original equations that contain the third variable.

6. Check the ordered triple in all three of the original equations.

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Example

Solution

Solve the system using elimination.3 9 6 3

2 2

2

x y z

x y z

x y z

The equations are in standard form.

(1)

(2)

(3)

2 2

2

x y z

x y z

(2)

(3)

(4)3x + 2y = 4 Adding

Eliminate z from equations (2) and (3).

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Eliminate z from equations (1) and (2).Multiplying equation (2) by 6

3 9 6 3

2 2

2

x y z

x y z

x y z

(1)

(2)

(3)

3 9 6 3

2 2

x y z

x y z

3 9 6 3

12 6 6 12

x y z

x y z

Adding15x + 15y = 15

Eliminate x from equations (4) and (5).

3x + 2y = 415x + 15y = 15

Multiplying top by 5

15x – 10y = 2015x + 15y = 15

Adding5y = 5y = 1

Using y = 1, find x from equation 4 by substituting.

3x + 2y = 43x + 2(1) = 4

x = 2

continued

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continued

Substitute x = 2 and y = 1 to find z.x + y + z = 22 – 1 + z = 2

1 + z = 2 z = 1

The solution is the ordered triple (2, 1, 1).

3 9 6 3

2 2

2

x y z

x y z

x y z

(1)

(2)

(3)

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Example

Solution

Solve the system using elimination.3 1

2 2 2

2 3 1

x y z

x y z

x y z

The equations are in standard form.

(1)

(2)

(3)

2 6 2 2

2 2 2

x y z

x y z

(1)

(2)

(4)5y 4z = 0 Adding

Eliminate x from equations (1) and (2).

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Eliminate x from equations (1) and (3).

3 1

2 2 2

2 3 1

x y z

x y z

x y z

(1)

(2)

(3)

3 1

2 3 1

x y z

x y z

5y + 4z = 2 (5)

Eliminate y from equations (4) and (5).

5y 4z = 05y + 4z = 2

0 = 2

All variables are eliminated and the resulting equation is false, which means that this system has no solution; it is inconsistent.

continued