Answer
Please see the graph.
Work Step by Step
The red line is the graph $ 3x-4y\le12$, and the blue line is the graph $ x^{2}+y^{2}<16$.
$3x-4y\ge 12$
Since we have a greater 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.
$(0,2)$
$3x-4y\le 12$
$3*0-4*2 \le 12$
$0-8 \le 12$
$ -8 \le 12$ (true, so we shade this side of the line)
$x^2+y^2 < 16$
Since we have a less than sign, we use a dotted line. We also pick the point $(0,2)$ to determine what side of the line to shade.
$(0,2)$
$x^2+y^2 < 16$
$0^2+2^2 < 16$
$0+4 < 16$
$4 < 16$ (true, so we shade the side of the graph with the point)
The overlap of the two graphs is the solution set.
