Chapter 4Section 4.6

Solving Systems of Linear Inequalities

Systems of Inequalities

A system of linear inequalities are several of linear inequalities all graphed on the same graph. The region where they all overlap is called the solution to the system.

To solve the system of inequalities above we do the following:

1. Graph each inequality putting arrows on the correct side of the line instead of shading.

2. Shade in the region they will all overlap.

3. Located the points where the lines cross by using either substitution or elimination methods for intersecting lines (if required)

8 7 6 5 4 3 2 1 1 2 3 4 5 6 7 8

8

7

6

5

4

3

2

1

1

2

3

4

5

6

7

8

Graph the system:

Solve the first inequality for y.

Slope -2Intercept 3

Solve the second inequality for y.

Slope Intercept -2

Graph the following system of inequalities and find the points where the borders cross at.

Solve the first inequality for x.

Vertical line at -3

Solve the second inequality for x.

Slope = Intercept = -3

8 7 6 5 4 3 2 1 1 2 3 4 5 6 7 8

8

7

6

5

4

3

2

1

1

2

3

4

5

6

7

8

The crossing point will be where the following lines intersect.

Substituting: The crossing point will have coordinates of:

4 3 2 1 1 2 3 4 5 6 7 8 9 10 11 12

4

3

2

1

1

2

3

4

5

6

7

8

9

10

11

12

Sometimes more than two inequalities will need to be graphed on the same graph.

Graph the system:

Graph:

Graph:

Graph:

Solve:

Graph this.

If you are careful enough about how you graph the lines you can read come of the crossing points right off the graph.

Cross: Cross:Find where

Cross at:

Find where

Cross at:

-6 -5 -4 -3 -2 -1 1 2 3 4 5 6

-6

-5

-4

-3

-2

-1

1

2

3

4

5

6

62

4

0

0

yx

y

y

x To solve the system of inequalities above we do the following:

1. Graph each inequality putting arrows on the correct side of the line instead of shading.

2. Shade in the region they will all overlap.

3. Located the points where the lines cross by using either substitution or elimination methods for intersecting lines

y = 0

x = 0

y = 4

y=-2x+6

Solve the inequality

y < -2x+6A

B

C

D

Point A is (0,0)

Point B is (0,4)

Point C is (3,0)

Point D is (1,4)

To find Point C

0

3

62

620

62

0

y

x

x

x

xy

y

To find Point D

4

1

22

624

62

4

y

x

x

x

xy

y The points A, B, C and D are called Corner Points.

-1 1 2 3 4 5 6

-1

1

2

3

4

5

6

Very often it is practical to have x and y not be negative and you get the inequalities x>0 and y>0 in the system of inequalities. Find the solution to the system below.

1125

52

0

0

yx

yx

y

x

x = 0

y = 0

Solve

52 xy

Solve

5.52

52

11

2

5

2

2

1152

xy

xy

xy

52 xy

5.525 xy

(0,0)

(0,5)

3

52

5)1(2

1

1125

1024

1125

522

1125

52

y

y

y

x

yx

yx

yx

yx

yx

yx

5

11

115

11)0(25

1125

0

x

x

x

yx

y

0,511

3,1

-1 1 2 3 4 5 6

-1

1

2

3

4

5

6

7

8

9

10

11

12

Solve:

1823

4

0

0

yx

x

y

x x=0

y=09

1832

1823

23

218

23

22

x

xy

y

xy

yx

x=4

923 xy

0,0

9,0

0,4

3

62

18212

182)4(3

1823

4

y

y

y

y

yx

x

3,4