How to simplify $42\sqrt{45} \over 7\sqrt{35}$? The problem is $$42\sqrt{45} \over 7\sqrt{35}$$ HELP! My daughter's math sheet shows how to reduce the squareroots, but the examples all use the same square-root; the problems show two different numbers square-rooted. Can you please help work this problem so I know how to help her with this assignment?
Thanks!
 A: Note that $45=5\cdot 9$, hence $\sqrt{45}=3\cdot \sqrt 5$. Also $\sqrt{35}=\sqrt 5\cdot\sqrt 7$. That should help simplify a lot.
A: Remember that for positive numbers $\sqrt{ab}=\sqrt{a}\times\sqrt{b}$. So we can simplify as follows:
$$\frac{42\times\sqrt{45}}{7\times\sqrt{35}}=\frac{6\times7\times\sqrt9\times\sqrt5}{7\times\sqrt7\times\sqrt5}=\frac{6\times 3}{\sqrt7}=\frac{18}{\sqrt7}=\frac{18}{\sqrt7}\times1=\frac{18}{\sqrt7}\times\frac{\sqrt7}{\sqrt7}=\frac{18\times\sqrt7}{7}$$
A: $$\frac{42\sqrt{45}}{7\sqrt{35}} = \frac{6\times 7\sqrt{9\times 5}}{7\sqrt{7\times 5}} $$ 
$$ = \frac{6\sqrt 9 \times \sqrt 5}{\sqrt 7 \times \sqrt 5} = \frac{6\times \sqrt 9}{\sqrt 7}$$
$$= \frac{6 \times 3}{\sqrt 7} \times \frac{\sqrt 7}{\sqrt 7} = \frac{18\sqrt 7}{ 7}$$ 
Cancel common factors, then multiply denominator to clear the square root.
A: $$\frac{42 \sqrt{45}}{7 \sqrt{35}} \;\; = \;\; \frac{42}{7} \cdot \frac{\sqrt{45}}{\sqrt{35}} \;\; = \;\; \frac{42}{7} \cdot \sqrt{\frac{45}{35}} \;\; = \;\; \frac{6}{1} \cdot \sqrt{\frac{9}{7}}$$
$$ = \;\; \frac{6}{1} \cdot \frac{\sqrt{9}}{\sqrt{7}} \;\; = \;\; \frac{6}{1} \cdot \frac{3}{\sqrt{7}} \;\; = \;\; \frac{6}{1} \cdot \frac{3}{\sqrt{7}} \cdot \frac{\sqrt{7}}{\sqrt{7}}$$
$$= \;\; \frac{6 \cdot 3 \cdot \sqrt{7}}{1 \cdot \sqrt{7} \cdot \sqrt{7}} \;\; =\;\; \frac{18\sqrt{7}}{7}$$
