Solution. Let $x$ be the numerator of the original fraction. Then $x+3$ is its denominator, and $\frac{x}{x+3}$ is the original fraction. After increasing the numerator and the denominator, we get the fraction $\frac{x+2}{3(x+3)}$. According to this, $\frac{x}{x+3}+\frac{x+2}{3(x+3)}=1$. If we extend the first fraction by 3 and add the resulting fractions sequentially, we get

$$
\begin{aligned}
& \frac{3 x}{3(x+3)}+\frac{x+2}{3(x+3)}=1 \\
& \frac{4 x+2}{3(x+3)}=1 \\
& 4 x+2=3 x+9 \\
& x=7
\end{aligned}
$$

Therefore, the original fraction is $\frac{x}{x+3}=\frac{7}{10}$.
Problem 2. In one fraction, the denominator is three times larger than the numerator. If we increase the numerator of this fraction by 2, and triple the denominator, then the sum of the thus obtained fraction and the original fraction is equal to 1. Determine the original fraction.