Solve:

x^2 + 2yz+ 5x = 2

y^2 + 2zx+ 5y = 2

z^2 + 2xy+ 5z = 2

Pen and paper only, please.

First subtract equation 2 from equation 1 to get:

x^2 - y^2 - 2xz + 2yz + 5x - 5y = 0

This factors into:

(x-y) * (x+y-2z+5) = 0

Therefore either one of the two factors equals 0:

__x=y OR x+y-2z+5=0__

The same procedure applied to equations 2 and 3 yields:

__y=z OR -2x+y+z+5=0__

Trying to apply the procedure to equations 1 and 3 at this point will result in redundant equations. A different trick is needed - add all three original equations together:

x^2 + y^2 + z^2 + 2xy + 2xz + 2yz + 5x + 5y + 5z = 6

The next step is not obvious - multiply both sides by 4 and then add 25 to both sides and factor to get:

(2x+2y+2z+5)^2 = 49

Then both sides are perfect squares and imply:

__2x+2y+2z+5 = 7 OR 2x+2y+2z+5 = -7__

Then there are eight possible systems of linear equations. Solving each of them yields all solutions to the original system: __(-2,-2,-2), (2,2,-3), (2,-3,2), (-3,2,2), (1/3,1/3,1/3), (-1/3,-1/3,-16/3), (-1/3,-16/3,-1/3), (-16/3,-1/3,-1/3)__.