To solve this system of equations, we can use substitution or elimination.

Using substitution, we can solve for one variable in terms of the other, then substitute that into the other equation to find the value of the variable.

From the second equation, we can solve for x in terms of y:
x = 3 - y

Now substitute x = 3 - y into the first equation:
(3 - y)^2 + y^2 = 29
Expanding and simplifying:
9 - 6y + y^2 + y^2 = 29
2y^2 - 6y - 20 = 0
Dividing by 2:
y^2 - 3y - 10 = 0
Factoring:
(y - 5)(y + 2) = 0
So, y = 5 or y = -2

Now plug these values back into x = 3 - y to find the corresponding values of x:
For y = 5:
x = 3 - 5
x = -2

For y = -2:
x = 3 - (-2)
x = 5

Therefore, the solutions to the system of equations are x = -2, y = 5 and x = 5, y = -2.