system of equation

Question

x^2 + y^2 + 2y = 32

x + y + 2xy = 22

find x and y :)
need clear solutions :)

Answer 1

x^2 + y^2 + 2y = 32

x + y + 2xy = 22

x(1+2y) = 22-y

x = (22-y)/(1+2y)

substitute into the 1st equation and solve for y

then plug that value into the last equation and solve for x

the 1st equation is a circle. Graph the 2nd equation and it's potentially intersecting the circle in 4 points.

[(22-y)/(1+2y)]^2 + y^2 +2y = 32

multiply by (1-2y)^2 to eliminate fractions. That creates a 4th degree equation with 4 potential solutions

(22-y)^2 + y2(1+2y)^2 + 2y(1+2y)^2 = 32(1+2y)^2

484 - 44y^2 + y^2(1+4y + 4y^2) + 2y(1+4y+4y^2) = 32(1+4y+4y^2)

combine like terms and use Descartes' Rule of Signs to determine maximum number of positive and negative real solutions. Imaginary solutions, if any, come in conjugate pairs.