We are given a System of Inequalities and are asked to graph the system, give two ordered pairs which are solutions, and two which are not.

{y \leq x + 3 y < - 3 (I) (II)

To draw a linear inequality, we must draw the boundary line and then shade the region of the coordinate plane containing the solutions. We will graph the inequalities one at a time.

Graphing Inequality (I)

Since the first inequality is

y \leq x + 3

, the equation for the boundary line is

y = x + 3 .

Notice that the equation is in slope-intercept form

y = m x + b,

where

m

is the slope and

b

is the

y -

intercept. By comparing them, we can find the parameters for the boundary line.

y = y = m x + b 1 x + 3

We can graph the boundary line by first plotting the

y -

intercept and then using the slope to find another point.

To determine if the solution set lies to the right or the left of the boundary line, we can use a test point. If a true statement is obtained after substituting the point into the inequality, we shade the region containing the point. Otherwise, we shade the opposite region. For simplicity, we will use

(0, 0)

as our test point.

y \leq x + 3

SubstituteII

$x = 0$ , $y = 0$

0 \leq ? 2 (0) + 3

ZeroPropMult

Zero Property of Multiplication

0 \leq 3

Since

0 \leq 3,

(0, 0)

is a solution to the inequality. Thus, we should shade the region to the right of the boundary line. Note that since the inequality symbol is

\leq,

the line will be solid.

Graphing Inequality (II)

We can graph the second inequality using a similar process. First, we will write the equation of the boundary line.

y = - 3

Notice that this time the boundary line is a horizontal line. To determine which region of the plane should be shaded, we will test

(0, 0)

again.

y < - 3

SubstituteII

$x = 0$ , $y = 0$

0 ≮ - 3

Since

0 ≮ - 3,

we will shade the region which does not contain

(0, 0) .

We can add the graph of Inequality (II) to the graph of Inequality (I).

The solution set of the system is represented by the overlapping shaded region.

Notice that there are infinitely many points inside and outside the solution region that we can choose from. For example, $(2, 4)$ and $(4, 4)$ are solutions, while $(2, 1)$ and $(4, 1)$ are not.

Exercise