$10 \cdot 7$ Let $x$ and $y$ be two natural numbers with two digits, and $x<y$. The product $xy$ is a four-digit number. The first digit is 2, and if this first digit 2 is removed, the remaining number is exactly $x+y$, for example $x=30, y=70$, besides this pair, there is another pair of numbers with the above property, try to find these two numbers.
(Dutch Mathematics Competition, 1983)
[Solution] From the given, we have $x y=2000+x+y$,
which means
$$
(x-1)(y-1)=2001=3 \cdot 23 \cdot 29 \text {. }
$$

Since $x$ and $y$ are both two-digit numbers, we have
$$
\begin{array}{l}
(x-1)(y-1)=29 \cdot 69, \\
(x-1)(y-1)=23 \cdot 87,
\end{array}
$$

which gives
$$
\left\{\begin{array} { l } 
{ x - 1 = 2 9 , } \\
{ y - 1 = 6 9 , }
\end{array} \quad \left\{\begin{array}{l}
x-1=23, \\
y-1=87 .
\end{array}\right.\right.
$$

Solving these, we get
$$
x=30, y=70 \text {, or } x=24, y=88 \text {. }
$$