Question

Solve for a and b, $\frac{\sqrt{2}+\sqrt{3}}{3\sqrt{2}-2\sqrt{3}}=a-b\sqrt{6}$. Find $6(a+b)$

Answer 1

$\frac{\sqrt{2}+\sqrt{3}}{3\sqrt{2}-2\sqrt{3}}$

Multiplying numerator and denominator by $3\sqrt{2}+2\sqrt{3}$

$\frac{\sqrt{2}+\sqrt{3}\times (3\sqrt{2}+2\sqrt{3})}{3\sqrt{2}-2\sqrt{3}\times (3\sqrt{2}+2\sqrt{3})}=\frac{6+2\sqrt{6}+3\sqrt{6}+6}{6}=\frac{12+5\sqrt{6}}{6}$

Hence $a=2$

$b=-\frac{5}{6}$

$a+b=2-\frac{5}{6}$

$a+b=\frac{7}{6}$

$6(a+b)=7$

Hence the answer is $7$