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

`x^2+y^2=29`

`x+y=3`

Notice that raising to square the sum x+y, yields:

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

You need to substitute 29 for `x^2+y^2` and 3 for `x+y ` such that:

`3^2 = 29 + 2xy`

`2xy = 9 - 29 =gt 2xy = -20 =gt xy = -10`

You need to write x in terms of y such that:

`x = 3-y`

You should substitute `3-y` for x in equation `xy = -10` such that:

`(3-y)y = -10`

`3y - y^2 + 10 = 0`

`y^2 - 3y - 10 = 0`

You need to use quadratic formula such that:

`y_(1,2) = (3+-sqrt(9+40))/2`

`y_(1,2) = (3+-7)/2`

`y_1 = 5 ; y_2 = -2`

You need to substitute 5 for y in `x = 3-y` such that:

`x = 3-5 =gt x = -2`

You need to substitute -2 for y in `x = 3-y` such that:

`x = 3-(-2) =gt x = 5`

**Hence, evaluating the solutions to simultaneous equations yields `x=-2;y=5` and `x=5,y=-2` .**