Question 1169062: Solve the system and graph the curves:
x^2-2x-4y^2+8y-2=0
5x^2-10x+12y^2+24y-58=0
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Answer by htmentor(1343)   (Show Source):
You can put this solution on YOUR website! If we multiply the 1st equation by 5 and subtract the 2nd from it, we have:
32y^2 - 16y - 48 = 0
Divide through by 16 and factor:
(2y - 3)(y + 1) = 0
y = 3/2, y = -1
Using the 1st equation to solve for x, we get:
x^2 - 2x - 4 - 8 - 2 = 0
x^2 - 2x - 14 = 0 -> x= -2.8729833 and x= 4.8729833
x^2 - 2x - 4(3/2)^2 + 8(3/2) - 2 = 0 -> x^2 - 2x + 1 = 0 -> (x-1)(x-1) = 0
This gives x = 1, multiplicity 2
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