A fraction is such that its denominator is 3 more than its numerator. When 1 is added to both the numerator and denominator, the result is $\dfrac{7}{8}$. Find the original fraction.

1. Define the unknown numerator and express the original fraction algebraically.

Let the numerator be $x$.

Since the denominator is 3 more than the numerator, the denominator is $x+3$.

Original fraction $=\dfrac{x}{x+3}$.

2. Form the equation after adding 1 to both numerator and denominator.

When 1 is added to both the numerator and denominator,

$\frac{x+1}{x+4}=\frac{7}{8}.$

3. Solve the fractional equation and find the original fraction.

$8(x+1)=7(x+4)$

$8x+8=7x+28$

$x=20$

So the numerator is $20$ and the denominator is $20+3=23$.

Original fraction $=\dfrac{20}{23}$.