SOLUTION: could you help me sketch the graph of the solution to: x^2+y^2<or=9 and y-2x>or=x-1. Also if you don't mind, please do the same with x>or=|y|, y>or=|-3|, 2y-x<or=4. In the first p

Question 64370: could you help me sketch the graph of the solution to: x^2+y^2or=x-1. Also if you don't mind, please do the same with x>or=|y|, y>or=|-3|, 2y-x
Answer by stanbon(75887) (Show Source): You can put this solution on YOUR website!
could you help me sketch the graph of the solution to: x^2+y^2or=x-1?
x^2+y^2=9 is a circle with center at (0.0) and a radius of 3.
x^2+y^2<9 is the interior of the circle.
y-2x=x-1 is the line y=-x-1 which passes through (0,-1) and (2,-3)
y-2x>x-1 the the half-plane above the line y=-x-1.
I can't graph the circle on this site but the line looks like this:

In the first problem I came up with a line passing through (0,-1) and (1,0)with shading above that line.
You are correct.
----------------------
Also if you don't mind, please do the same with x>or=|y|, y>or=|-3|, 2y-x In the second one I came up with a line through (-3), a line through (0,0) and (2,-2); and a v opening right from (0,0)going through (2,2)(2,-2).
That is the equality graph. The greater than part is the inside of the
V-shape.
Cheers,
Stan H.
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