Posted by Saud on Wednesday, November 21, 2012 at 11:17am.
just substitute the linear into the quadratic:
y = 2-3x
2x^2 + 3x(2-3x) + (2-3x)^2 = 0
2x^2 + 6x - 9x^2 + 4 - 12x + 9x^2 = 0
2x^2 - 6x + 4 = 0
2(x-1)(x-2) = 0
solutions: (1,-1)(2,-4)
or,
2x^2 + 3xy + y^2 = 0
(2x+y)(x+y) = 0
y = -2x or y = -x
3x+(-2x) = 2
x = 2, so y = -4
3x+(-x) = 2
x = 1, so y = -1
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