2. (3 points) There are already several red and black balls in the box. After adding some black balls, the red balls account for one quarter of the total number of balls; after adding some red balls, the number of red balls is two-thirds of the black balls. If the number of black balls and red balls added is the same, then the original ratio of red balls to black balls in the box is $\qquad$ .
【Analysis】We set the number of red balls to be $a$, and the number of black balls to be $b$, and the number of black and red balls added to be $x$. According to the conditions “after adding some black balls, the red balls account for one quarter of all the balls; after adding some red balls, the number of red balls is two-thirds of the black balls. If the number of black balls and red balls added is the same,” we can set up two equations to solve.

【Solution】Solution: Let the number of red balls be $a$, the number of black balls be $b$, and the number of black and red balls added be $x$.
$$
\begin{array}{c}
\frac{a}{x+a+b}=\frac{1}{4}, \\
x+a+b=4a, \\
x=3a-b, \\
\frac{a+x}{b+x}=\frac{2}{3}, \\
3a+3x=2b+2x, \\
x=2b-3a,
\end{array}
$$

Substitute $x=3a-b$ into the equation,
$$
3a-b=2b-3a,
$$
$$
\begin{array}{l}
3b=6a, \\
a: b=1: 2,
\end{array}
$$

The original ratio of red balls to black balls in the box is 1:2.
The answer is: 1: 2.