Exercise 3 Page 257

Graphing a single inequality involves two main steps.

1. Plotting the boundary line.
2. Shading half of the plane to show the solution set.

Here, we need to do this process for each of the inequalities in the system. The system's solution set will be the intersection of the shaded regions in the graphs of (I) and (II).

Boundary Lines

We can tell a lot of information about the boundary lines from the inequalities given in the system.

• Exchanging the $\text{inequality}$ $\text{symbols}$ for $\text{equals}$ $\text{signs}$ gives us the boundary line equations.
• Observing the $\text{inequality}$ $\text{symbols}$ tells us whether the inequalities are strict.
• Writing the equation in slope-intercept form will help us highlight the slopes $m$ and $y\text{-}$intercepts $b$ of the boundary lines.

Let's find each of these key pieces of information for the inequalities in the system.

Information	Inequality (I)	Inequality (II)
Given Inequality	$y\ge \text{-}x+4$	$x+y\le 0$
Boundary Line Equation	$y=\text{-}x+4$	$x+y=0$
Solid or Dashed?	$\ge \Rightarrow$ Solid	$\le \Rightarrow$ Solid
$y=mx+b$	$y=\text{-}1x+4$	$y=\text{-}1x+0$

Great! With all of this information, we can plot the boundary lines.

Shading the Solution Sets

Before we can shade the solution set for each inequality, we need to determine on which side of the plane their solution sets lie. To do that, we will need a test point that does not lie on either boundary line.

It looks like the point $(1,0)$ would be a good test point. We will substitute this point for $x$ and $y$ in the given inequalities and simplify. If the substitution creates a true statement, we shade the same region as the test point. Otherwise, we shade the opposite region.

Information	Inequality (I)	Inequality (II)
Given Inequality	$y\ge \text{-}x+4$	$x+y\le 0$
Substitute $(1,0)$	$(0)\stackrel{?}{\ge }\text{-}(1)+4$	$(1)+(0)\stackrel{?}{\le }0$
Simplify	$0\ngeqq 3\times$	$1\nleqq 0\times$
Shaded Region	opposite	opposite

For both the inequalities, we will shade the region opposite our test point. For Inequality (I), it will be above the boundary line. For Inequality (II), however, it will be below the boundary line.

Now that we have graphed the system, we see that there is no overlapping region. This means that there are no points that are a solution to the system, only points that are solutions to each individual inequality.