SOLUTION: 2x+y=3 x^2+y^2+4y=6

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Question 869409: 2x+y=3
x^2+y^2+4y=6
Answer by Fombitz(32388) About Me  (Show Source):
You can put this solution on YOUR website!
1.2x%2By=3
2.x%5E2%2By%5E2%2B4y=6
From eq. 1,
y=-2x%2B3
Substitute into eq. 2,
x%5E2%2B%28-2x%2B3%29%5E2%2B4%28-2x%2B3%29=6
x%5E2%2B%284x%5E2-12x%2B9%29-8x%2B12=6
5x%5E2-20x-15=0
x%5E2-4x-3=0
%28x-3%29%28x-1%29=0
Two solutions,
x-3=0
x=3
Then
y=-2%283%29%2B3
y=-6%2B3
y=-3
and
x-1=0
x=1
Then
y=-2%281%29%2B3
y=1
(3,-3) and (1,1)
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