If $\frac{5+3 \sqrt{3}}{7+4 \sqrt{3}}=a+b \sqrt{3},$ find the value of $a$ and $b$.

(A) To find the values of $a$ and $b$,we rationalize the denominator of the given expression:
$\frac{5+3 \sqrt{3}}{7+4 \sqrt{3}} \times \frac{7-4 \sqrt{3}}{7-4 \sqrt{3}}$
$= \frac{(5+3 \sqrt{3})(7-4 \sqrt{3})}{(7)^2 - (4 \sqrt{3})^2}$
$= \frac{35 - 20 \sqrt{3} + 21 \sqrt{3} - 12(3)}{49 - 16(3)}$
$= \frac{35 + \sqrt{3} - 36}{49 - 48}$
$= \frac{-1 + \sqrt{3}}{1}$
$= -1 + 1 \sqrt{3}$
Comparing this with $a+b \sqrt{3}$,we get $a = -1$ and $b = 1$.