Solution:
The answer is $f(x) = c x + a + b$, $g(x) = c x + a$, $h(x) = c x + b$, where $a$, $b$, $c$ are real numbers. Obviously these solutions work, so we wish to show they are the only ones.

First, put $y = 0$ to get $f(x+0) = g(x) + h(0)$, so $g(x) = f(x) - h(0)$. Similarly, $h(y) = f(y) - g(0)$. Therefore, the functional equation boils down to $f(x+y) = f(x) + f(y) - (g(0) + h(0))$. By shifting and appealing to Cauchy's functional equation (with $f$ continuous) we get $f(x) = c x + g(0) + h(0)$, $g(x) = c x + g(0)$ and $h(x) = c x + h(0)$.