Answer
Please see the graph.
Work Step by Step
The red graph is $ y\le-x^{2}+3$, and the blue graph is $ y\le2x-1$.
$ y\le-x^{2}+3$
Since we have a less than or equal to sign, we use a solid line. We also pick the point $(0,2)$ to determine what side of the line to shade.
$(0,2)$
$ y\le-x^{2}+3$
$2\le-0^{2}+3$
$2 \le 0+3$
$2 \le 3$ (true, so we shade the side of the graph with the point)
$ y\le2x-1$
Since we have a less than or equal to sign, we use a solid line. We also pick the point $(0,2)$ to determine what side of the line to shade.
$(0,2)$
$ y\le2x-1$
$2\le 2*0-1$
$2 \le 0-1$
$2 \le -1$ (false, so we shade the side of the graph without the point)
The overlap of the two graphs is the solution set.
