#### Answer

See graph.

#### Work Step by Step

A line in slope-intercept form has the equation:
$y=mx+b$ ($m=slope$, $b=y-intercept$)
We put our first equation into slope-intercept form:
$3x-y\geq 3$
$-y\geq -3x+3$
$y\leq 3x-3$
Thus we graph a line with $slope=3$ and $y-intercept=-3$.
We know that if an inequality has an "equal" sign (e.g. "$\leq$" or "$\geq$"), then a solid line should be graphed. If the inequality does not have an "equal" sign (e.g. "$\lt$" or "$\gt$"), then a dashed line should be graphed.
We also know that if an inequality has a "less than" sign (e.g. "$y\lt$..." or "$y\leq$..." or "$x\lt$..." or "$x\leq$..."), then the shading should be below the line (or left for vertical lines). If the inequality has a "greater than" sign (e.g. "$y\gt$..." or "$y\geq$..." or "$x\gt$..." or "$x\geq$..."), then the shading should be above the line (or right for vertical lines).
In our case, we must graph a solid line and shade below.
Next, we look at our second equation:
$y\lt 3$
This has the form of a horizontal line ($y=constant$) going through $y=3$.
We see from the inequality that we must graph a dashed line and shade below.
See the resulting graph. The dark region is the area where both inequalities overlap (e.g. the solution region).