Let's denote the full capacity of the tank as $C$ litres.

Initially, the tank is at $\frac{1}{4}$ of its full capacity. After adding 3 litres, the tank is at $\frac{1}{3}$ of its full capacity.

The fuel already in the tank at the start is $\frac{1}{4} C$. After adding 3 litres, the total fuel in the tank is $\frac{1}{4} C + 3$. This is now equal to $\frac{1}{3} C$.

We can now set up the equation: $\frac{1}{4} C + 3 = \frac{1}{3} C$

To solve for $C$, subtract $\frac{1}{4} C$ from both sides: $3 = \frac{1}{3} C - \frac{1}{4} C$

To simplify the right-hand side, find a common denominator: $\frac{1}{3} C - \frac{1}{4} C = \frac{4}{12} C - \frac{3}{12} C = \frac{1}{12} C$

Now the equation becomes: $3 = \frac{1}{12} C$

Multiply both sides by 12: $C = 36$

Therefore, the full capacity of the tank is $36$ litres.

Would you like more details on any step or have any questions?

Here are some related questions to explore:

1. How would the problem change if the initial fuel level was different?
2. What if instead of 3 litres, 5 litres were added? How would that affect the result?
3. Can you create a general formula for solving similar problems involving fuel capacity?
4. How does this problem illustrate the use of fractions in algebra?
5. Could this method be used to determine other quantities in real-life scenarios?

Tip: When solving word problems, carefully translate the given conditions into mathematical equations, as this often simplifies the problem.