14. [9] Real numbers $x$ and $y$ satisfy the following equations:
$$
\begin{array}{l}
x=\log _{10}\left(10^{y-1}+1\right)-1 \\
y=\log _{10}\left(10^{x}+1\right)-1
\end{array}
$$

Compute $10^{x-y}$.
Answer: $\frac{101}{110}$

Solution: Taking 10 to the power of both sides in each equation, these equations become:
$$
\begin{array}{l}
10^{x}=\left(10^{y-1}+1\right) \cdot 10^{-1} \\
10^{y}=\left(10^{x}+1\right) \cdot 10^{-1} .
\end{array}
$$

Let $a=10^{x}$ and $b=10^{y}$. Our equations become:
$$
\begin{array}{l}
10 a=b / 10+1 \\
10 b=a+1
\end{array}
$$
and we are asked to compute $a / b$. Subtracting the equations gives
$$
10 a-10 b=b / 10-a \Longrightarrow 11 a=101 b / 10,
$$
giving an answer of $101 / 110$.