[Algebra]
Problem: At present, the sum of the ages of a husband and wife, $P$, is six times the sum of the ages of their children, $C$. Two years ago, the sum of the ages of the husband and wife was ten times the sum of the ages of the same children. Six years from now, it will be three times the sum of the ages of the same children. Determine the number of children.

Solution: Let $n$ be the number of children.

At the present, $P=6 C$, where $P$ and $C$ are as given.  (1)

Two years ago, the sum of the ages of the husband and wife was $P-4$, since they were each two years younger.

Similarly, the sum of the ages of the children was $C-n(2)$ ( $n$ is the number of children).

So two years ago, $P-4=10(C-2 n)$    (2) , from the given condition.

Similarly, six years from now, $P+12=3(C+6 n)$       (3), from the given condition.

We want to solve for $n$.

Substituting (1) into each of (2) and (3),

$$
\begin{aligned}
& 6 C-4=10(C-2 n) \quad \text { or } \quad 20 n-4 C=4 \quad \text { or } \quad 5 n-C=1 \\
& 6 C+12=3(C+6 n) \quad \text { or } \quad-18 n+3 C=-12 \quad \text { or } \quad-6 n+C=-4
\end{aligned}
$$

Adding these two equations, $-n=-3$, so $n=3$.

Therefore, there were three children.

Final answer: 3