Answer
Please see the graph.
Work Step by Step
The red line is the graph $ 4x+3y\ge12$, and the blue line is the graph $ x^{2}+y^{2}<16$.
$4x+3y\ge 12$
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.
$(0,2)$
$4x+3y\ge 12$
$4*0+3*2 \ge 12$
$0+6 \ge 12$
$ 6 \ge 12$ (false, so we shade the other 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.
