Answer
Please see the graph.
Work Step by Step
Red line: $x^{2}-y^{2}\ge1$
Blue line: $\frac{x^{2}}{16}+\frac{y^{2}}{4}\le1$
Green line: $y\ge1$
$x^{2}-y^{2}\ge1$
This is the equation for a hyperbola. We have three regions to test--keeping $y$ the same for all three tests: $(-∞, -1)$, $(-1, 1)$, and $(1, ∞)$.
We pick the points $(-2, 0)$, $(0,0)$, and $(2,0)$ to determine what region(s) to shade.
$(-2,0)$
$x^{2}-y^{2}\ge1$
$(-2)^{2}-0^{2}\ge1$
$4 - 0 \ge 1$
$4 \ge 1$ (true, so we shade this region)
$(0,0)$
$x^{2}-y^{2}\ge1$
$0^{2}-0^{2}\ge1$
$0 - 0 \ge 1$
$0 \ge 1$ (false)
$(2,0)$
$x^{2}-y^{2}\ge1$
$(2)^{2}-0^{2}\ge1$
$4 - 0 \ge 1$
$4 \ge 1$ (true, so we shade this region)
$\frac{x^{2}}{16}+\frac{y^{2}}{4}\le1$
This is the equation for an ellipse. If the point is part of the solution set, then the point is in the ellipse. We pick the point $(0,0)$ to determine if it is inside the ellipse or outside the ellipse.
$\frac{x^{2}}{16}+\frac{y^{2}}{4}\le1$
$\frac{0^{2}}{16}+\frac{0^{2}}{4}\le1$
$\frac{0}{16}+\frac{0}{4}\le1$
$0 + 0 \le 1$
$0 \le 1$ (the point is inside the ellipse, so we shade inside the ellipse)
$y\ge1$
We pick the point $(0,0)$ to determine what side of the line to shade.
$y\ge1$
$0 \ge 1$ (false, so we shade the side of the line without the point)
The solution set of the set of inequalities is the overlap of the three graphs.
