Question

graph each system of inequalities by hand. 36. $\begin{cases}x^{2}+y^{2}geq9\\x + yleq3end{cases}$

Explanation:

Step1: Analyze the first - inequality

The inequality $x^{2}+y^{2}\geq9$ represents the region outside and on the circle centered at the origin $(0,0)$ with radius $r = 3$, since the equation of a circle is $(x - a)^{2}+(y - b)^{2}=r^{2}$, here $a = 0$, $b = 0$ and $r = 3$.

Step2: Analyze the second - inequality

Rewrite the inequality $x + y\leq3$ as $y\leq - x+3$. The line $y=-x + 3$ has a $y$ - intercept of 3 and a slope of - 1. The region represented by $y\leq - x + 3$ is the region below and on the line $y=-x + 3$.

Step3: Find the intersection region

The solution of the system of inequalities is the intersection of the region outside and on the circle $x^{2}+y^{2}=9$ and the region below and on the line $y=-x + 3$.

To graph by hand:

1. Draw the circle $x^{2}+y^{2}=9$ as a solid circle (because of $\geq$) and shade the region outside the circle.
2. Draw the line $y=-x + 3$ as a solid line (because of $\leq$) and shade the region below the line. The overlapping shaded region is the solution of the system of inequalities.

Answer:

Graph the circle $x^{2}+y^{2}=9$ (solid) and shade outside it, graph the line $y=-x + 3$ (solid) and shade below it. The intersection of the two shaded regions is the solution set.