Example 1 (1979 Yunnan Province Mathematics Competition Question) A four-digit number, its unit digit and hundred digit are the same. If the digits of this four-digit number are reversed (i.e., the thousand digit and unit digit are swapped, the hundred digit and ten digit are swapped), the new number minus the original number equals 7812. Find the original four-digit number.


Let the thousands, hundreds, and tens digits of the number be $x, y, z$, respectively, then
the original number $=10^{3} \cdot x+10^{2} \cdot y+10 \cdot z+y$,
the new number after reversing $=10^{3} \cdot y+10^{2} \cdot z+10 y+x$.
(2)-(1) gives $7812=999(y-x)+90(z-y)$,

which simplifies to $868=111(y-x)+10(z-y)$
$$
=10^{2} \cdot(y-x)+10 \cdot(z-x)+(y-x) .
$$

Comparing the hundreds, tens, and units digits on both sides of (3) gives
$$
y-x=8, z-x=6 \text {. }
$$

The thousands digit $x$ of the original number cannot be zero, so $x \geqslant 1$. Since $y=x+8 \geqslant 9$, and it is evident that the hundreds digit $y \leqslant 9$, we have $y=9, x=1, z=x+6=7$.
Therefore, the required four-digit number is 1979.