數學(聯立方程)

1) 2x-3y=2
x = (3y + 2)/2
So
x^2+y^2+3x-2=0 becomes
[(3y+2)/2]^2 + y^2 + 3(3y+2)/2 - 2 = 0
[(3y+2)^2]/4 + y^2 + (9y + 6)/2 - 2 = 0
(3y+2)^2 + 4y^2 + 2(9y + 6) - 8 = 0
9y^2 + 12y + 4 + 4y^2 + 18y + 12 - 8 = 0
13y^2 + 30y + 8 = 0
(13y + 4)(y + 2) = 0
y = - 4/13 or - 2
x = (3(- 4/13) + 2)/2 = 7/13 or (3(-2) + 2)/2 = - 2
(x , y) = (7/13 , - 4/13)
or
(x , y) = ( - 2 , - 2)