Surds question grade 9

Question

I am a student and I need help answering this question.

Simplify:

$\frac{6}{\sqrt{28}}$-$\frac{9}{\sqrt{63}}$

What I did:

$\frac{6}{\sqrt{28}}$- $\frac{9}{\sqrt{63}}$

=$\frac{6}{\sqrt{7×4}}$ - $\frac{9}{\sqrt{7×4}}$

= $\frac{6}{2\sqrt{7}}$ - $\frac{9}{3\sqrt{7}} $

$\frac{6}{2\sqrt{7}}$ × $\frac{2\sqrt{7}}{2\sqrt{7}}$ = $\frac{12\sqrt{7}}{14}$

$\frac{9}{3\sqrt{7}}$ × $\frac{3\sqrt{7}}{3\sqrt{7}}$ = $\frac{27\sqrt{7}}{63}$

($\frac{12\sqrt{7}}{14}$)/2 = $\frac{6\sqrt{7}}{7}$

($\frac{27\sqrt{7}}{63}$) /2 = $\frac{3\sqrt{7}}{7}$

$\frac{6\sqrt{7}}{7}$ - $\frac{3\sqrt{7}}{7}$

= $\frac{3\sqrt{7}}{7}$

Thankyou and help is appreciated.

Answer 1

Simply note that $$\frac{6}{\sqrt{28}}-\frac{9}{\sqrt{63}}=\frac{6}{2\sqrt7}-\frac{9}{3\sqrt7}=\frac{3}{\sqrt7}-\frac{3}{\sqrt7}=0$$

Answer 2

Hint: $\frac{6}{\sqrt{28}} = \frac{3}{\sqrt{7}} = \frac{9}{\sqrt{63}}$

You made some typos in your work.