Answer
See below
Work Step by Step
Given: $3y^2+x^2+4x+18y=-28\\9y^2-4x^2+8x+90y=-185$
Solve for y.
$$3y^2+x^2+4x+18y=-28\\(3y^2+18y)+x^2+4x=-28\\3(y+3)^2=-x^2-4x-1\\(y+3)^2=\frac{1}{3}(-x^2-4x-1)\\y+3=\pm \sqrt \frac{1}{3}(-x^2-4x-1)\\y=\pm \sqrt \frac{1}{3}(-x^2-4x-1)-3$$
$$9y^2-4x^2+8x+90y=-185\\(9y^2+90y)-4x^2+8x=-185\\9(y+5)^2=4x^2-8x+40\\(y+5)^2=\frac{1}{9}(4x^2-8x+40)\\y+4=\pm \sqrt \frac{1}{9}(4x^2-8x+40)\\y=\pm \frac{1}{3}\sqrt 4x^2-8x+40-5$$
The solutions are $(-2.319,-2.017);(-0.296,-2.821)$
