# How do you solve -2r<= -6 and graph the solutions?

Dec 20, 2016

The shaded region in the graph

#### Explanation:

Use $r = \sqrt{{x}^{2} + {y}^{2}} \ge 0$ and change of sign, in an inequality,

reverses the the inequality operator.

$I f - a \le - b$, then $a > b$

Converting to cartesian form,

$- \sqrt{{x}^{2} + {y}^{2}} \le - 3 \to \sqrt{{x}^{2} + {y}^{2}} > 3 \to {x}^{2} + {y}^{2} > 9$

This represents the region exterior to the circle ${x}^{2} + {y}^{2} = 0$.

graph{x^2+y^2>9 [-10, 10, -5, 5]}