Example 1. There is a four-digit number. It is known that its tens digit minus 1 equals the units digit, and its units digit plus 2 equals the hundreds digit. The sum of this four-digit number and the number formed by reversing the order of its four digits equals 9878. Try to find this four-digit number.
Solution: Let the required four-digit number be $a b c d$. According to the problem, we have $\overline{a b c \bar{d}}+\overline{d c b a}=9878$,

which means $\left(10^{3} \cdot a+10^{2} \cdot b+10 \cdot c+d\right)+\left(10^{3} \cdot d\right.$ $\left.+10^{2} \cdot c+10 \cdot b+a\right)=9878$, or $10^{3} \cdot(a+d)+10^{2} \cdot(b+c)+10 \cdot(b+c)$ $+(a+d)=9878$.
By comparing the first and last two digits on both sides of (1), we get
$$
a+d=8 \text {. }
$$

Thus, $b+c=17$.
Also, $c-1=d, d+2=b$,

which means $b-c=1$,
Therefore, the solution is $a=1, b=9, c=8, d=7$.
Hence, the required four-digit number is 1987.
In the solution process, the equations (4) and (5) were not substituted into (1) to solve the indeterminate equation, but instead, the equations (3) and (4) were directly obtained by comparing the first and last two digits of (1). This avoids complex discussions. However, this technique has certain limitations and should not be misused.