# Given A = {(x,y) belong in RXR/x^2+y^2-2xy=1; x+y+xy=5} and S = Sigma(x+y) (sum) , what is S?

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### 1 Answer

The problem provides the information that the elements of the set A are solutions to the simultaneous equations, such that:

`{(x^2 + y^2 - 2xy = 1),(x + y + xy = 5):}`

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

`{((x+y)^2 - 2xy - 2xy = 1),(x + y + xy = 5):}`

You need to come up with the following substitutions, such that:

`x+y = p`

`xy = q`

Replacing `p` for `x+y` and `q` for `xy` yields:

`{(p^2 - 4q = 1),(p + q = 5):} => {(p^2 - 4q = 1),(p = 5 - q):}`

Replacing `5 - q` for `p` in the top equation yields:

`{((5 - q)^2 - 4q = 1),(p = 5 - q):}`

`(5 - q)^2 - 4q = 1 => 25 - 10q + q^2 - 4q - 1 = 0`

`q^2 - 14q + 24 = 0`

You need to use quadratic formula such that:

`q_(1,2) = (14+-sqrt(196 - 96))/2 => q_(1,2) = (14+-sqrt100)/2`

`q_(1,2) = (14+-10)/2 => q_1 = 12 ; q_2 = 2`

`p_1 = 5 - q_1 => p_1 = 5 - 12 => p_1 = -7`

`p_2 = 5 - q_2 => p_2 = 5 - 2 => p_2 = 3`

Replacing back `x+y` for `p` and `xy` for `q` yields:

`{(x+y = -7),(xy = 12):} `

` `

You need to use Lagrange's resolvents such that:

`t^2 -pt + q = 0 => t^2 + 7t + 12 = 0`

`t_(1,2) = (-7 +- sqrt(49 - 48))/2 => t_(1,2) = (-7 +- 1)/2`

`t_1 = -3 ; t_2 = -4 => {(x = -3),(y = -4):} or {(x = -4),(y = -3):}`

`{(x+y = 3),(xy = 2):} `

Using Lagrange's resolvents yields:

`t^2 - pt + q = 0 => t^2 -3t + 2 = 0 `

`t_(1,2) = (3+-sqrt(9-8))/2 => t_(1,2) = (3+-1)/2`

`t_1 = 2; t_2 = 1`

`{(x = 1),(y = 2):} or {(x = 2),(y = 1):}`

You need to evaluate the following summation `Sigma(x+y)` , such that:

`Sigma(x+y)` =` -3 - 4 - 4 - 3 + 1 + 2 + 2 + 1`

`Sigma(x+y) = -8`

**Hence, evaluating the summation `Sigma(x+y)` , under the given conditions, yields **`Sigma(x+y) = -8.`

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