How to solve the system x+y=3 and x^2/y+y^2/x=9/2?
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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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This is a symmetrical system and we'll solve it using the sum and the product.
x + y = S (sum)
x*y = P (product)
We'll multiply the 2nd equation by xy both sides:
x^3 + y^3 = 9xy/2
We'll re-write the equations of the system:
x + y = 3
x^3 + y^3 = 9xy/2
We'll write the sum of cubes from the second equation:
x^3 +y^3 = (x+y)(x^2 + xy + y^2)
Now, we'll substitute all by S and P.
S = 3
3(S^2 - 2P) = 9P/2
S^2 - 2P = 3P/2
9 - 2P = 3P/2
18 - 6P = 3P
9P = 18
P = 2
We'll form the quadratic equation knowing the sum and the product:
x^2 - 3x + 2 = 0
x = 1 and y = 2
x = 2 and y = 1
The solutions of the system are (1 ; 2) and (2 ; 1).
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