Answer
Please see the graph.
Work Step by Step
Red line: $x^{2}-y^{2}<1$
Blue line: $\frac{x^{2}}{16}+y^{2\ }\le1$
Green line: $x\ge-2$
$x^{2}-y^{2}<1$
This is the equation of a hyperbola, and we have three regions to test (keeping the $y$ value the same): $(-∞, -1)$, $(-1, 1)$, and $(1, ∞)$.
We pick the x-values $-2$, $0$, and $2$ (all with a y-value of $0$) to test.
$(-2,0)$
$x^{2}-y^{2}<1$
$(-2)^{2}-0^{2}<1$
$4 - 0 < 1$
$4 < 1$ (false)
$(0,0)$
$x^{2}-y^{2}<1$
$0^{2}-0^{2}<1$
$0-0 <1$
$0 < 1$ (true, so we shade the side of the line with the point)
$(2,0)$
$x^{2}-y^{2}<1$
$(2)^{2}-0^{2}<1$
$4 - 0 < 1$
$4 < 1$ (false)
$\frac{x^{2}}{16}+y^{2\ }\le1$
This is the equation of an ellipse. We pick the point $(2,2)$ to determine whether to shade inside the ellipse or outside the ellipse.
$\frac{2^{2}}{16}+2^{2}\le1$
$\frac{4}{16}+4\le1$
$1/4 + 4 \le 1$
$17/4 \le 1$ (false, so we shade inside the ellipse)
$x\ge-2$
We pick the point $(2,2)$ to determine what side of the line to shade.
$x\ge-2$
$2\ge-2$ (true, so we shade the side of the line with the point)
The solution set to the set of inequalities is the overlap of all three graphs.
