To express \(\frac{7 - 4\sqrt{3}}{7 + 4\sqrt{3}}\) in the form \(a + b\sqrt{3}\), we can multiply the numerator and the denominator by the conjugate of the denominator. The conjugate of \(7 + 4\sqrt{3}\) is \(7 - 4\sqrt{3}\).
Let's perform the calculation:
\[ \frac{7 - 4\sqrt{3}}{7 + 4\sqrt{3}} \cdot \frac{7 - 4\sqrt{3}}{7 - 4\sqrt{3}} = \frac{(7 - 4\sqrt{3})^2}{(7 + 4\sqrt{3})(7 - 4\sqrt{3})} \]
Calculating the denominator:
\[ (7 + 4\sqrt{3})(7 - 4\sqrt{3}) = 7^2 - (4\sqrt{3})^2 = 49 - 48 = 1 \]
Now calculating the numerator:
\[ (7 - 4\sqrt{3})^2 = 7^2 - 2 \cdot 7 \cdot 4\sqrt{3} + (4\sqrt{3})^2 = 49 - 56\sqrt{3} + 48 = 97 - 56\sqrt{3} \]
Thus, we have:
\[ \frac{7 - 4\sqrt{3}}{7 + 4\sqrt{3}} = \frac{97 - 56\sqrt{3}}{1} = 97 - 56\sqrt{3} \]
This can be expressed as:
\[ a + b\sqrt{3} \quad \text{where} \quad a = 97 \quad \text{and} \quad b = -56 \]
Now we need to find \(a + b\):
\[ a + b = 97 - 56 = 41 \]
Thus, the final answer is:
\[ \boxed{41} \]