SOLUTION: Solve the system: { x^2 + y^2 = 50 { 2x^2 - 3xy + 3y^2 = 50

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Question 1101022: Solve the system:
{ x^2 + y^2 = 50
{ 2x^2 - 3xy + 3y^2 = 50
Answer by Alan3354(69443) About Me  (Show Source):
You can put this solution on YOUR website!
{ x^2 + y^2 = 50
{ 2x^2 - 3xy + 3y^2 = 50
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y^2 = 50-x^2
y = sqrt(50-x^2)
Sub for y & y^2
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2x^2 + 3(50-x^2) - 3x*sqrt(50-x^2) = 50
2x^2 + 150 - 3x^2 - 3x*sqrt(50-x^2) = 50
x^2 + 3x*sqrt(50-x^2) = 100
x^2 - 100 = 3x*sqrt(50-x^2)
x^4 - 200x^2 + 10000 = 450x^2 - 9x^4
10x^4 - 650x^2 + 10000 = 0
x^4 - 65x^2 + 1000 = 0
(x^2 - 25)*(x^2 - 40) = 0
x = +/-5 --> y = +/-5
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x = +/-2sqrt(10) --> y = +/-sqrt(10)
Check for extraneous solutions.