For three numbers $a, b$, and $c$, the following equations hold:

$$
a+b=332, a+c=408 \text{ and } b+c=466 \text{. }
$$

Determine the numbers $a, b$, and $c$.
Solution. We add the three equations and get

$$
2(a+b+c)=1206, \text { i.e. } a+b+c=603 \text {. }
$$

If we sequentially substitute $a+b=332$, $a+c=408$, and $b+c=466$ into the last equation, we get

$$
332+c=603, 408+b=603 \text { and } a+466=603
$$

from which we find $c=271, b=195$ and $a=137$.