Answer
$ x^{2}+y^{2} \leq 25$
$x+2y \geq 5 $
.
Work Step by Step
First sentence: $\quad x^{2}+y^{2} \leq 25 $
Second sentence: $\quad x+2y \geq 5 $
1st inequality ($ \leq $ inequality sign, border solid)::
$x^{2}+y^{2} = 25$ is a circle, centered at (0,0), radius 5.
Test point: (0,0). $\quad $Is $ 0^{2}+0^{2} \leq 25 \quad ?$
Yes. shade the region containing (0,0).
2nd inequality ($ \geq$ inequality sign, border solid):
Graph $x+2y = 5:$
$2y=-x+5$
$y=-\displaystyle \frac{1}{2}x+\frac{5}{2}$
y intercept=$\displaystyle \frac{5}{2}$, and for x=5, y=$0$ .
Draw the line through (0,$\displaystyle \frac{5}{2}$) and ($5,0$)
Test point: (0,0). $\quad $Is $0+2(0) \geq 5 ?$
No. shade the region not containing (0,0).
The solution set is the region with both shadings.
(grey on the screenshot)
