Let the original fraction be $\frac{x - 3}{x}$, where $x$ represents the denominator. According to the problem, if we increase both the numerator and denominator by 1, the fraction becomes $\frac{3}{4}$. Therefore, the equation is:

$\frac{(x - 3) + 1}{x + 1} = \frac{3}{4}$

Simplify the numerator on the left-hand side:

$\frac{x - 2}{x + 1} = \frac{3}{4}$

Now, we can cross-multiply to solve for $x$:

$4(x - 2) = 3(x + 1)$

Expanding both sides:

$4x - 8 = 3x + 3$

Now, subtract $3x$ from both sides:

$x - 8 = 3$

Add 8 to both sides:

$x = 11$

Now that we know the denominator $x = 11$, the numerator is:

$x - 3 = 11 - 3 = 8$

Thus, the original fraction is:

$\frac{8}{11}$

Would you like more details on this solution or have any questions?

Here are 5 related questions to deepen your understanding:

1. How would the solution change if the fraction became $\frac{2}{3}$ instead of ( \frac{3}{