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) About Me  (Show Source):
You can put this solution on YOUR website!
1.2x%5E2-y%5E2-4+=0
2.x%5E2%2B3y%5E2=23
Multiply eq. 2 by -2 and add the equations together,
2x%5E2-y%5E2-4-2x%5E2-6y%5E2=0-46
-7y%5E2-4=-46
-7y%5E2=-42
y%5E2=6
y=0+%2B-+sqrt%286%29
From eq. 1,
2x%5E2-6-4=0
2x%5E2=10
x%5E2=5
x=0+%2B-+sqrt%285%29
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( +-sqrt%285%29+ ,-sqrt%286%29)
(sqrt%285%29,-sqrt%286%29)
(-sqrt%285%29,sqrt%286%29)
(sqrt%285%29,sqrt%286%29)
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