1

Warm Up1. Solve and graph |x – 4| < 2

2. Solve and graph |2x – 3| > 12x – 3 < -1 or 2x – 3 > 1

+3 +3 +3 +3

2x < 2 2x > 4

x – 4 > -2 and x – 4 < 2

+4 +4 +4 +4

x > 2 and x < 6

-6 -4 -2 0 2 4 6 8-8

x < 1 or x > 2

-6 -4 -2 0 2 4 6 8-8

2

Inequalities in Two VariablesThe solutions of an inequality in two variables are the ordered pairs of numbers that make the inequality true.To check whether an ordered pair is a solution of an inequality in two variables, substitute into the inequality and check for truth.

Example 1: Determine whether (2, 1) is a solution of x + y < 4

Substitute 2 for x and 1 for y2 + 1 < 43 < 4 TRUE(2, 1) is a solution of the inequality x + y < 4

3

Inequalities in Two VariablesYou try it! Determine whether the ordered pair is a solution to the inequality.

Example 2: (5, -3) and 2x – y > 5

Example 3: (4, 8) and y > 2x + 1

2(5) – (-3) > 5

10 + 3 > 5

13 > 5 TRUE, it is a solution

8 > 2(4) + 1

8 > 9 FALSE, not a solution

4

2. The set of points above the line. This is the half-plane where y > 2x + 2

y > 2x + 2

3. The set of points below the line. This is the half-plane where y < 2x + 2

Graphing Inequalities in Two VariablesThe graph of the linear equation y = 2x + 2separates the coordinate plane into three sets:

x-4 -3 -2 2 3 4 5-5 -1 1

y

-4-3-2

1234

-5

5

-1

y < 2x + 2

y =

2x +

21. The set of points on the boundary line (where y = 2x + 2.)

5

Now, test a point not on the line to determine which half-plane is the solution set.

Testing (0, 0) yields: 0 < 4 TRUE

So, the solution lies on that side of the line. Shade it in.

Graphing Inequalities in Two VariablesThe solutions of an inequality in two variables are all the points in a half-plane and may include the points on the boundary line.

x-4 -3 -2 2 3 4 5-5 -1 1

y

-4-3-2

1234

-5

5

-1

x – y < 4

Example 1: Graph x – y < 4

You can use any method to graph the boundary line x – y = 4.

The intercepts are: (0, -4) and (4, 0)

Graph with a dashed line to show that the boundary line is not included in the solution set.

6

Testing (0, 0) yields 0 12 TRUE

So, the solution lies on that side of the line. Shade it in.

Graphing Inequalities in Two VariablesYou try it! Graph 2x + 3y 12

x-4 -3 -2 2 3 4 5-5 -1 1

y

-4-3-2

1234

-5

5

-1

2x + 3y 12

The boundary line is:2x + 3y = 12

The x-intercept is (6, 0) and the y-intercept is (0, 4).

Graph with a solid line to show that the boundary line is included in the solution set.

7

Testing (0, 0) yields 0 4 TRUE

So, the solution lies on that side of the line. Shade it in.

Graphing Inequalities in Two VariablesYou try it! Graph x 4 on the plane.

x-4 -3 -2 2 3 4 5-5 -1 1

y

-4-3-2

1234

-5

5

-1

x 4

The boundary line is x = 4

Graph with a solid line to show that the boundary line is included in the solution set.

8

Testing (1, 1) yields:1 – 3 > 0 or –1 > 0 FALSE

So, the solution lies on the other side of the line. Shade it in.

Graphing Inequalities in Two VariablesYou try it! Graph y – 3x > 0

x-4 -3 -2 2 3 4 5-5 -1 1

y

-4-3-2

1234

-5

5

-1

y – 3x > 0

The boundary line is:y – 3x = 0

Solving for y yields y = 3x

The boundary line has a slope of 3 and a y-intercept of (0,0).

Graph with a dashed line to show that the boundary line is not included in the solution set.