Beginning Algebra, 11th Edition

NOTE Alternatively in Example 1,we can find the required region by solving the

given inequality for y.

Inequality from Example 1

Subtract 2x.

Divide by 3.

Ordered pairs in which yis equal to are on the boundary line, so pairs in

which y is less than will be belowthat line. As we move downvertically, the

y-values decrease.) This gives the same region that we shaded in FIGURE 36. (Ordered

pairs in which yis greater than will be abovethe boundary line.)

Graphing a Linear Inequality

Graph

This inequality does notinclude the equals symbol. Therefore, the points on the

line do notbelong to the graph. However, the line still serves as a bound-

ary for two regions, one of which satisfies the inequality.

To graph the inequality, first graph the equation Use a dashed lineto

show that the points on the line are notsolutions of the inequality See

FIGURE 37.

Now choose a test point to see which side of the line satisfies the inequality.

Original inequality

Let and

0 75 False

0 - 07 x= 0 y= 0.

?

5

x-y 75

x-y 7 5.

x-y=5.

x- y= 5

x- y 7 5.

EXAMPLE 2

- 23 x+ 2

- 23 x+ 2

- 23 x+ 2

y...-

2

3

x+ 2

3 y...- 2 x+ 6

2 x+ 3 y... 6

SECTION 3.5 Graphing Linear Inequalities in Two Variables 225

is a

convenient

test point.

1 0, 0 2

Since is false, the graph of the inequality is the region that does notcontain

1 0, 0 2 .Shade the otherregion, as shown in FIGURE 37, to obtain the required graph.

07 5

x

y

(0, 0)

(4, –3)

0

5

–5

x – y = 5

x – y > 5

FIGURE 37

NOW TRY
EXERCISE 2
Graph. 2 x- 4 y 78

NOW TRY ANSWER
2.

x

y

(^0) –2 4
2 x – 4y > 8

To check that the correct region is shaded, we test a point in the shaded region.

For example, use from the shaded region as follows.

CHECK Original inequality

Let and

✓ True

This true statement verifies that the correct region was shaded in FIGURE 37.

NOW TRY

7 7 5

4 - 1 - 327 x= 4 y=-3.

?

5

x-y 7 5

1 4, - 32

Use parentheses

to avoid errors.