x^2+2xy=16 , 3x^2-6=4xy-2y^2
amitnrw:
x = 2 y = 3 or x = -2 y = - 3
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so what do wanna em get here
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Given : x²+2xy=16 , 3x²-6=4xy-2y²
To find : Integral Solution
Solution:
x²+2xy=16 ,
3x²-6=4xy-2y²
x²+2xy=16 ,
=> 2xy=16 - x²
3x²-6=4xy-2y²
=>3x²- 6 = 2(2xy) - 2y²
=>3x²- 6 = 2(16 - x²) - 2y²
=> 3x²- 6 = 32 - 2x² - 2y²
=> 5x² + 2y² = 38
Lets check integral Values
x = 1 => y² = 33/2
x = 2 => y² = 9 => y = ± 3
x = 3 or more not Possible
as x² is there hence
Possible solutions
x = 2 , y = 3
x = - 2 & y = - 3
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