# solve system in real x y xy+x+y=47 x^2+y^2=74

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### 1 Answer

You should come up with the following notations for x+y and xy such that:

`x+y=p`

`xy=q`

You should substitute p for x+y and q for xy in the system of simultaneous equations such that:

`q+p = 47`

`p^2 - 2q = 74`

Notice that `p^2-2q` substitutes`x^2+y^2` such that:

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

You should substitute p for s+y and q for xy such that:

`p^2 = x^2 + 2q + y^2`

`x^2 +y^2 = p^2 - 2q`

Hence, you need to solve for p and q the system of equations such that:

`q+p = 47 =gt q = 47-p`

`p^2 - 2(47-p) = 74`

You need to open the brackets in the second equation such that:

`p^2 - 94 + 2p - 74 = 0`

`p^2 + 2p - 168 = 0`

`p_(1,2) = (-2+-sqrt(4+672))/2`

`p_(1,2) = (-2+-sqrt676)/2`

`p_(1,2) = (-2+-26)/2`

`p_1 =12 ; p_2 = -14`

`q = 47-p =gt q_1 = 47-12 = 35`

`q = 47-p =gt q_2 = 47 + 14 = 61`

You need to solve for x and y the equations:

`x^2 - 12x + 35 = 0`

`x^2 + 14x + 61 = 0`

`x_(1,2) = (12+-sqrt(144 - 140))/2`

`x_(1,2) = (12+-2)/2`

`x_1 = 7 ; x_2 = 5`

`x^2 + 14x + 61 = 0`

`x_(1,2) = (-14+-sqrt(-48))/2`

Since `sqrt(-48) !in R` , there are no real solutions to the equation `x^2 + 14x + 61 = 0` .

**Hence, evaluating the solutions to the system of equations yields x = 7 and y = 5 ; x = 5 and y = 7.**

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