Exercise 30 Page 162

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	$2x+6y>12$	$3x+9y\le 27$
Boundary Line Equation	$2x+6y=12$	$3x+9y=27$
Solid or Dashed?	$>\Rightarrow$ Dashed	$\le \Rightarrow$ Solid
$y=mx+b$	$y=\text{-}\frac{1}{3}x+2$	$y=\text{-}\frac{1}{3}x+3$

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 $(0,0)$ 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	$2x+6y>12$	$3x+9y\le 27$
Substitute $(0,0)$	$2(0)+6(0)\stackrel{?}{>}12$	$3(0)+9(0)\stackrel{?}{\le }27$
Simplify	$0\ngtr 12$	$0\le 27$
Shaded Region	opposite	same

For Inequality (I) we will shade the region opposite our test point, or above the boundary line. For Inequality (II), however, we will shade the region containing the test point, or below the boundary line.

Finally, we can view only the solution set by removing the shaded regions that are not overlapping.