How to solve the system x+y=3 and x^2/y+y^2/x=9/2?
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Tushar Chandra
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We have to solve the system x + y = 3 and x^2/y + y^2/x = 9/2
x + y = 3
=> x = 3 - y
Substitute this in x^2/y + y^2/x = 9/2
=> (3 - y)^2 / y + y^2 / (3 - y) = 9/2
=> (3 - y)(3 - y)^2 + y* y^2 / (3 - y) = (9/2)*y*(3 - y)
=> (3 - y)^3 + y^3 = (9/2)(3y - y^2)
=> 2(3^3 - 3*9*y + 3*3*y^2 - y^3 + y^3) = 27y - 9y^2
=> 54 - 81y + 27y^2 = 0
=> y^2 - 3y + 2 = 0
=>y^2 - 2y - y + 2 = 0
=>y( y - 2) - 1( y - 2) = 0
=> (y - 1)(y - 2) = 0
=> y = 1 and y =2
=> x = 3 - y = 2 and 1.
The values of x and y are (2, 1) and (1, 2)
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