# Solve the system of equations. x^2/y+y^2/x=35/3 x+y=10

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### 2 Answers

You need to solve for x and y the system of simultaneous equations, such that:

`{(x^2/y + y^2/x = 35/3),(x + y = 10):}`

`{((x^3 + y^3)/(xy) = 35/3),(x + y = 10):}`

Converting the sum of cubes into a product yields:

`x^3 + y^3 = (x + y)(x^2 - xy + y^2)`

`{((x + y)(x^2 - xy + y^2)/(xy) = 35/3),(x + y = 10):}`

Substituting 10 for `x + y` yields:

`{((10)(x^2 - xy + y^2)/(xy) = 35/3),(x + y = 10):}`

Reducing duplicate factors yields:

`{((x^2 - xy + y^2)/(xy) = 7/6 ),(x + y = 10):}`

Using Vieta's relations yields:

`x^2 + y^2 = (x + y)^2 - 2xy`

Substituting `(x + y)^2 - 2xy` for `x^2 + y^2` yields:

`{((100 - 2xy - xy)/(xy) = 7/6),(x + y = 10):}`

`{((100 - 3xy)/(xy) = 7/6),(x + y = 10):}`

`{(100/(xy) - 3 = 7/6),(x + y = 10):}`

`{(100/(xy) = 7/6 + 3),(x + y = 10):}`

`{(100/(xy) = 25/6),(x + y = 10):} => {(4/(xy) = 1/6),(x + y = 10):} `

`{(xy = 24),(x + y = 10):}`

Using Vieta's relations yields:

`x + y = -b/a = 10`

`xy = c/a = 24`

`x^2 - (b/a)x + c/a = 0 => x^2 - 10x + 24 = 0`

Completing the square `x^2 - 10x` yields:

`x^2 - 10x+ 25 = -24 + 25 => (x - 5)^2 = 1 => x - 5 = +-1`

`x_1 = 5 + 1 => x_1 = 6`

`x_2 = 5 - 1 => x_2 = 4`

**Hence, evaluating the solutions to the system yields `x = 6, y = 4` and `x = 4, y = 6` .**

We'll multiply the first equation by 3xy:

3(x^3 + y^3) = 35xy

But x^3 + y^3 = (x+y)(x^2 - xy + y^2)

3(x+y)(x^2 - xy + y^2) = 35xy

We'll substitute x+y by the second equation:

x+y=10

30(x^2 - xy + y^2) = 35xy

x^2 + y^2 = (x+y)^2 - 2xy

x^2 + y^2 = 100 - 2xy

30(100 - 3xy) = 35xy

3000 - 90xy = 35xy

125xy = 3000

xy = 24

We'll form the quadratic equation when knowing the sum and the product:

x^2 - Sx + P = 0

x^2 - 10x + 24 = 0

x1 = [10+sqrt(100 - 96)]/2

x1 = (10+2)/2

x1 = 6

x2 = (10-2)/2

x2 = 4

**So, the solutions of the symmetric system are: {4 ; 6} and {6 ; 4}.**