If both $a$ and $b$ are rational numbers, find the values of $a$ and $b$ in each following equalities
$\frac{3 + \sqrt{7}}{3 - \sqrt{7}} = a + b \sqrt{7}$
$\frac{5 + \sqrt{3}}{7 - 4 \sqrt{3}} = 47 a + b \sqrt{3}$

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Solution

(a)

Rationalizing the LHS of $\frac{3 + \sqrt{7}}{3 - \sqrt{7}} = a + b \sqrt{7}$ .

$\frac{3 + \sqrt{7}}{3 - \sqrt{7}} \times \frac{3 + \sqrt{7}}{3 + \sqrt{7}} = a + b \sqrt{7}$

$\frac{{(3 + \sqrt{7})}^{2}}{{(3)}^{2} - {(\sqrt{7})}^{2}} = a + b \sqrt{7}$

$\frac{9 + 7 + 6 \sqrt{7}}{9 - 7} = a + b \sqrt{7}$

$\frac{16 + 6 \sqrt{7}}{2} = a + b \sqrt{7}$

$8 + 3 \sqrt{7} = a + b \sqrt{7}$

By comparing both sides,

$a = 8, b = 3$

(b)

Rationalizing the LHS of $\frac{5 + \sqrt{3}}{7 - 4 \sqrt{3}} = 47 a + b \sqrt{3}$ .

$\frac{5 + \sqrt{3}}{7 - 4 \sqrt{3}} = 47 a + b \sqrt{3}$

$\frac{5 + \sqrt{3}}{7 - 4 \sqrt{3}} \times \frac{7 + 4 \sqrt{3}}{7 + 4 \sqrt{3}} = 47 a + b \sqrt{3}$

$\frac{(5 + \sqrt{3}) (7 + 4 \sqrt{3})}{{(7)}^{2} - {(4 \sqrt{3})}^{2}} = 47 a + b \sqrt{3}$

$\frac{35 + 20 \sqrt{3} + 7 \sqrt{3} + 12}{49 - 48} = 47 a + b \sqrt{3}$

$47 + 27 \sqrt{3} = 47 a + b \sqrt{3}$

By comparing both sides,

$a = 1, b = 27$

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