Exercise 13 Page 642

We want to solve the given system of inequalities by graphing. Let's graph each of inequality one at a time.

Inequality (I)

We can write the boundary curve for Inequality (I) by replacing the less than sign with an equals sign. The above equation is an equation of an ellipse. Let's write it in a standard form. The denominator of the $x\text{-}$variable is greater than the denominator of the $y\text{-}$variable. Therefore, we are given the equation of a horizontal ellipse. Let's find necessary information to draw a graph.

	Horizontal Ellipse
Standard-Form Equation	$\frac{(x-h{)}^2}{a^2}+\frac{(y-k{)}^2}{b^2}=1$	$\frac{(x-0{)}^2}{{\left(4\sqrt{2}\right)}^2}+\frac{(y-0{)}^2}{2^2}=1$
Center	$(h,k)$	$(0,0)$
Vertices	$(h\pm a,k)$	$(0\pm 4\sqrt{2},0)$ $\Updownarrow$ $(\text{-}4\sqrt{2},0)$ and $(4\sqrt{2},0)$
Co-vertices	$(h,k\pm b)$	$(0,0\pm 2)$ $\Updownarrow$ $(0,\text{-}2)$ and $(0,2)$
$a,b,c$ relationship, $a>b>0$	$c^2=a^2-b^2$	$c^2={\left(4\sqrt{2}\right)}^2-2^2$
Foci	$(h\pm c,k)$	$(0\pm c,0)$

Let's calculate the value of $c$ and find the foci. Since we will add and subtract $c$ to find the foci, we kept the principal root when calculating its value. We will now use the information we found to draw the ellipse. Since we have a strict inequality, the boundary curve will be dashed. Now that we have the boundary curve, we need to determine which region to shade. To do so, we will use $(0,0)$ as a test point. If the point satisfies the inequality, we will shade the region that contains the point. If not, we will shade the opposite region. Since the substitution produced a true statement, we will shade the region that contains the point $(0,0).$

Inequality (II)

We can write the boundary curve for Inequality (II) by replacing the less than sign with an equals sign. The graph of this function is the graph of $y=|x|$ after a few transformations. Let's consider some transformations.

Transformations of $y=|x|$
Vertical Translations	$$\begin{array}{c}\text{Translation up }k\text{ units, }k>0\\ y=|x|+k\end{array}$$
	$$\begin{array}{c}\text{Translation down }k\text{ units, }k>0\\ y=|x|-k\end{array}$$
Horizontal Translations	$$\begin{array}{c}\text{Translation right }h\text{ units, }h>0\\ y=|x-h|\end{array}$$
	$$\begin{array}{c}\text{Translation left }h\text{ units, }h>0\\ y=|x+h|\end{array}$$
Reflections	$$\begin{array}{c}\text{In the }x\text{-axis}\\ y=\text{-}|x|\end{array}$$
	$$\begin{array}{c}\text{In the }y\text{-axis}\\ y=|\text{-}x|\end{array}$$

Using the table, we can see that there is a horizontal translation right by $2$ units, reflection in the $x\text{-}$axis and a vertical translation up by $2$ units. Applying the transformations to $y=|x|,$ we can draw the boundary line. Because we have a strict inequality, the boundary line will be dashed. Once again, we will check which region we should shade. To do so, we will use $(1,0)$ as a test point. Since the substitution produced a true statement, we will shade the region which contains the point $(1,0).$

Solution

The solution set is the overlapping region.