Question

In the system shown below, what are the coordinates of the solution that lies in quadrant IV?
2x^2+y^2=33
x^2+y^2+2y=19

Answer 1

2x^2 + y^2 = 33
x^2 + y^2 + 2y = 19

2x^2 + y^2 = 33
-2(x^2 + y^2 + 2y = 19)

2x^2 + y^2 = 33
-2x^2 - 2y^2 - 4y = -38

-y^2 - 4y = -5

-y^2 - 4y + 5 = 0

y^2 + 4y - 5 = 0

(y + 5)(y - 1) = 0

y = -5, 1

In quadrant IV, the y is negative. Therefore, 1 is not an answer.

y = -5

2x^2 + y^2 = 33

2x^2 + (-5)^2 = 33

2x^2 + 25 = 33

2x^2 = 8

x^2 = 4

x = ±4
In quadrant IV, the x is positive. Therefore, -4 is not an answer.

x = 4

Therefore, the solution is Quadrant IV is (4, -5)

Answer 2

it was (2,-5) -_-