In each of the following determine rational numbers $a$ and $b$:$ \frac{5+3 \sqrt{3}}{7+4 \sqrt{3}}=a+b \sqrt{3} $

Given:

\( \frac{5+3 \sqrt{3}}{7+4 \sqrt{3}}=a+b \sqrt{3} \)

To do: 

We have to determine rational numbers $a$ and $b$.

Solution:

We know that,

Rationalising factor of a fraction with denominator ${\sqrt{a}}$ is ${\sqrt{a}}$.

Rationalising factor of a fraction with denominator ${\sqrt{a}-\sqrt{b}}$ is ${\sqrt{a}+\sqrt{b}}$.

Rationalising factor of a fraction with denominator ${\sqrt{a}+\sqrt{b}}$ is ${\sqrt{a}-\sqrt{b}}$.

LHS $=\frac{5+3 \sqrt{3}}{7+4 \sqrt{3}}=\frac{(5+3 \sqrt{3})(7-4 \sqrt{3})}{(7+4 \sqrt{3})(7-4 \sqrt{3})}$

$=\frac{35-20 \sqrt{3}+21 \sqrt{3}-12 \times 3}{(7)^{2}-(4 \sqrt{3})^{2}}$

$=\frac{35+\sqrt{3}-36}{49-48}$

$=\frac{\sqrt{3}-1}{1}$

$=-1+\sqrt{3}$

Therefore,

$a+b \sqrt{3}=-1+\sqrt{3}$

Comparing both sides, we get,

$a=-1$ and $b=1$

Hence, $a=-1$ and $b=1$.