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Solving Systems of Equations

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Solving Systems of Equations

Graphing Linear Inequalities

- How do we graph an inequality
- Define a boundary line
- Graphing a boundary line
- Define the solution for a system of inequalities
- Find the solution of a system of inequalities

- Solution of an inequality are all the ordered pairs (points) that make the inequality true.

Graph

Consider the inequality

y ≥ x

y = x

REMEMBER: Solution are all

the ordered pairs (points) that

make the inequality true.

6

5

4

3

Boundary line

2

1

1

2

3

4

5

6

Consider the inequality

y ≥ x

Pick two points from each

side of the graph

6

5

4

(1,3)

3

2

(4,1)

1

1

2

3

4

5

6

substitute into

Consider the inequality

y ≥ x

Check points if they make

inequality true.

(1,3)

y ≥ x

6

5

4

(1,3)

3

2

(4,1)

1

1

2

3

4

5

6

Consider the inequality

y ≥ x

Check points if they make

inequality true.

(1,3)

y ≥ x

6

substitute into

3 ≥ 1

5

4

(1,3)

3

2

(4,1)

1

1

2

3

4

5

6

Consider the inequality

y ≥ x

Check points if they make

inequality true.

(1,3)

y ≥ x

6

substitute into

3 ≥ 1

5

4

(1,3)

3

2

(4,1)

1

1

2

3

4

5

6

substitute into

Consider the inequality

y ≥ x

Check points if they make

inequality true.

(4,1)

y ≥ x

6

5

4

(1,3)

3

2

(4,1)

1

1

2

3

4

5

6

Consider the inequality

y ≥ x

Check points if they make

inequality true.

(4,1)

y ≥ x

6

substitute into

1 ≥ 4

5

4

(1,3)

3

2

(4,1)

1

1

2

3

4

5

6

Consider the inequality

y ≥ x

Check points if they make

inequality true.

(4,1)

y ≥ x

6

substitute into

1 ≥ 4

X

5

4

(1,3)

3

2

X

(4,1)

1

1

2

3

4

5

6

Consider the inequality

y ≥ x

Shade the side where the correct

point lies.

6

5

4

(1,3)

3

2

X

(4,1)

1

1

2

3

4

5

6

Consider the inequality

y ≥ x

Shade the side where the correct

point lies.

6

5

4

(1,3)

3

2

1

1

2

3

4

5

6

1

y = x - 2

2

Consider the inequality

x - 2y ≤ 4

x - 2y = 4

Graph

3

2

1

1

2

3

4

5

6

-1

-2

1

y = x - 2

2

(0,1)

(6,0)

Consider the inequality

x - 2y ≤ 4

x - 2y = 4

Graph

¡¡TEST POINTS !!

3

2

1

1

2

3

4

5

6

-1

-2

substitute into

Consider the inequality

x - 2y ≤ 4

(0,1)

x - 2y ≤ 4

3

2

(0,1)

1

(6,0)

1

2

3

4

5

6

-1

-2

Consider the inequality

x - 2y ≤ 4

(0,1)

x - 2y ≤ 4

substitute into

0 - 2(1) ≤ 4

-2 ≤ 4

3

2

(0,1)

1

(6,0)

1

2

3

4

5

6

-1

-2

Consider the inequality

x - 2y ≤ 4

(6,0)

x - 2y ≤ 4

substitute into

3

2

(0,1)

1

(6,0)

1

2

3

4

5

6

-1

-2

Consider the inequality

x - 2y ≤ 4

(6,0)

x - 2y ≤ 4

substitute into

6 - 2(0) ≤ 4

X

6 ≤ 4

3

2

(0,1)

1

(6,0)

X

1

2

3

4

5

6

-1

-2

Consider the inequality

x - 2y ≤ 4

¡¡ SHADE CORRECT REGION !!

3

2

(0,1)

1

(6,0)

X

1

2

3

4

5

6

-1

-2

2

GRAPH

y = x + 3

3

6

5

4

3

2

1

1

2

3

4

5

6

1.

3y - 2x ≥ 9

(0,5)

6

5

4

3

(3,0)

2

1

1

2

3

4

5

6

1.

3y - 2x ≥ 9

2

GRAPH

y = x + 3

3

TEST!!

(0, 5)

3(5) - 2(0) ≥ 9

15 - 0 ≥ 9

X

(0,5)

1

y = x + 1

3

6

Graph

5

4

3

2

1

1

2

3

4

5

6

2.

x - 3y > -3

TEST!!

(0, 5)

X

0 - 3(5) > -3

0 - 15 > -3

X

3

2

1

-3

-2

-1

1

2

3

-1

Consider the system

x + y ≥ -1

-2x + y < 2

Graph

3

2

(0,0)

1

-3

-2

-1

1

2

3

-1

Consider the system

x + y ≥ -1

-2x + y < 2

y = - x - 1

TEST: (0,0)

0 + 0 ≥ -1

0 ≥ -1

Graph

3

2

1

-3

-2

-1

1

2

3

(0,0)

-1

Consider the system

x + y ≥ -1

-2x + y < 2

y = 2x + 2

TEST: (0,0)

-2(0) + 0 < 2

0 < 2

3

3

2

2

1

1

-3

-3

-2

-2

-1

-1

1

1

2

2

3

3

-1

-1

x + y ≥ -1

-2x + y < 2

3

2

1

-3

-2

-1

1

2

3

-1

Consider the system

x + y ≥ -1

-2x + y < 2

SOLUTION:

- Lies where the two shaded
- regions intersect each
- other.

Graph

3

2

y = x - 2

1

(0,0)

-2

-1

1

2

3

4

2

3

Consider the system

-2x + 3y < -6

5x + 4y < 12

X

TEST: (0,0)

-2(0) + 3(0) < -6

-1

0 < -6

X

-2

Graph

3

2

y = - x + 3

1

(0,0)

-2

-1

1

2

3

4

5

4

Consider the system

-2x + 3y < -6

5x + 4y < 12

TEST: (0,0)

5(0) + 4(0) < 12

-1

0 < 12

-2

Graph

3

2

1

(0,0)

-2

-1

1

2

3

4

Consider the system

-2x + 3y < -6

5x + 4y < 12

SOLUTION:

- Lies where the two shaded
- regions intersect each
- other.

-1

-2

Graph

3

2

1

(0,0)

-2

-1

1

2

3

4

Consider the system

-2x + 3y < -6

5x + 4y < 12

NOTE:

- All order pairs in dark
- region are true in both
- inequalities.

-1

-2

Graph

6

4

2

(0,0)

2

4

6

8

10

12

-2

-4

-6

Consider the system

x - 4y ≤ 12

4y + x ≤ 12

TEST: (0,0)

(0) - 4(0) ≤ 12

0 - 0 ≤ 12

0 ≤ 12

Graph

6

4

2

(0,0)

2

4

6

8

10

12

-2

-4

-6

Consider the system

x - 4y ≤ 12

4y + x ≤ 12

TEST: (0,0)

4(0) + (0) ≤ 12

0 ≤ 12