solve the system (x^2/16)+(y^2/4) ≤1 and y>(1/2)x-2?

Answer 1

Part one
(x^2/16)+(y^2/4) ≤1

4 *[ (x^2/16)+(y^2/4) ≤1]
4* (x^2/16)+ 4* (y^2/4) ≤ 4* 1
(x^2)/4 + y^2 ≤4
(x^2)/4 ≤ 4 - y^2
4*[ (x^2)/4 ≤ 4 - y^2]
x^2 ≤ 16 - 4y^2
x^2 + 4y^2 -16 ≤ 0


Part two

y>(1/2)x-2
2* [y>(1/2)x-2]
2y > x - 4
2y - x + 4 > 0

Answer 2

I just sketched the graph of this, and I see the answer.

(x^2)/16 + (y^2)/4 = 1

is an ellipse with horizontal major axis 8 and vertical minor axis 4. It's centered at the origin and passes through these points: (0,2), (-4,0), (0,-2), and (4,0). The direction of the inequality indicates we're interested in the interior of the ellipse, including the boundary line.

The line y = x/2 - 2 is a straight line with positive slope 1/2 and y-intercept (0,-2). We note immediately that this line passes through the point (4,0). The direction of the inequality indicates we're interested in the region to the right of the line, not including the line itself.

The soluion to your problem is the area in the fourth quadrant bounded by the ellipse (including the boundary line itself) and the line (but not including the line itself). The endpoints (0,-2) and (4,0) are not part of the solution,

The shape of the solution set roughly resembles a crescent moon.

You could figure out the enclosed area by first solving the ellipse equation for y (bottom half only), then subtracting the equation of the line, and integrating the resulting expression on x from 0 to 4.

But I don't think that was part of your question.

Answer 3

using www.quickmath.com

Click on Plot under Inequalities, then click Advanced, then type in

((x^2)/16) + ((y^2)/4) <= 1
y > (1/2)x - 2

Click Tick Marks and set the margins to
-10,10
-10,10

then click pot, and you should get an ellipse from (-4,0) to (4,0), and (0,2) to (0,-2), and the inside of the ellipse is shaded. The only thing about it is, that there is a diagonal line from (0,-2) to (4,0)