Solve the system of equations: 2x + 2y = 80 and 30x + 30y = 40.

We have:
Equation 1': $x + y = 40$
Equation 2': $3x + 3y = 4$

Notice that the left side of Equation 2' can be factored:

From Equation 1', we know that $x + y = 40$. Substitute this into $3(x + y) = 4$:

The statement $120 = 4$ is false. This means that there are no values of $x$ and $y$ that can satisfy both equations simultaneously. The lines represented by these equations are parallel and distinct, so they never intersect.

The system of equations has $\boxed{\text{no solution}}$.