SOLUTION: Solve the system for real values of x and y
2x^2-y^2-4 =0
x^2+3y^2=23
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Question 330301: Solve the system for real values of x and y
2x^2-y^2-4 =0
x^2+3y^2=23
Answer by Fombitz(32388) (Show Source): You can put this solution on YOUR website!
1.
2.
Multiply eq. 2 by -2 and add the equations together,
From eq. 1,
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