Answer
Please see the graph.
Work Step by Step
Red line: $3x-4y\le12$
Blue line: $x^{2}+y^{2}<16$
$3x-4y\le12$
We pick a point to determine which side of the line to shade. We pick $(0,0)$.
$(0,0)$
$3x-4y\le12$
$3*0-4*0\le12$
$0 - 0 \le 12$
$0 \le 12$ (true, so we shade the side of the line with this point)
$x^{2}+y^{2}<16$
The related equation $x^{2}+y^{2}=16$ is the equation of a circle with radius 4. If the point is inside the circle, then it is part of the solution set.
The solution set of the two equations is the overlapping area. $(0,0)$ is a part of the solution set.
