# 004 Sample Final A, Problem 4

Graph the system of inequalities. ${\displaystyle y>2x-3\qquad y\leq 4-x^{2}}$ Solution:

Step 1:
First we replace the inequalities with equality. So ${\displaystyle y=\vert x\vert +1}$, and ${\displaystyle x^{2}+y^{2}=9}$.
Now we graph both functions.
Step 2:
Now that we have graphed both functions we need to know which region to shade with respect to each graph.
To do this we pick a point an equation and a point not on the graph of that equation. We then check if the
point satisfies the inequality or not. For both equations we will pick the origin.
${\displaystyle y<\vert x\vert +1:}$ Plugging in the origin we get, ${\displaystyle 0<\vert 0\vert +1=1}$. Since the inequality is satisfied shade the side of
${\displaystyle y<\vert x\vert +1}$ that includes the origin. We make the graph of ${\displaystyle y<\vert x\vert +1}$, since the inequality is strict.
${\displaystyle x^{2}+y^{2}\leq 9:}$ ${\displaystyle (0)^{2}+(0)^{2}=0\leq 9}$. Once again the inequality is satisfied. So we shade the inside of the circle.
We also shade the boundary of the circle since the inequality is ${\displaystyle \leq }$
The final solution is the portion of the graph that below ${\displaystyle y=\vert x\vert +1}$ and inside ${\displaystyle x^{2}+y^{2}=9}$