# How do you simplify (2sqrt(7)+35)/[sqrt(7)]?

Nov 4, 2015

$2 + 5 \sqrt{7}$

#### Explanation:

It would help if there was something that we could cancel out. At first glance, there is a $\sqrt{7}$ in the numerator and the denominator, so lets see if we can do something about that. If we factor $35$ we get;

$35 = 5 \cdot 7$

But the $7$ can be further factored by taking the square root.

$7 = {\sqrt{7}}^{2} = \sqrt{7} \cdot \sqrt{7}$

So $35$ becomes;

$35 = 5 \cdot \sqrt{7} \cdot \sqrt{7}$

Now we can start simplifying the expression.

$\frac{2 \sqrt{7} + 35}{\sqrt{7}} = \frac{2 \sqrt{7} + 5 \cdot \sqrt{7} \cdot \sqrt{7}}{\sqrt{7}}$

Move the $\sqrt{7}$ out of the numerator.

$\frac{\sqrt{7} \left(2 + 5 \sqrt{7}\right)}{\sqrt{7}}$

Now the $\sqrt{7}$s cancel and we have;

$\frac{\cancel{\sqrt{7}} \left(2 + 5 \sqrt{7}\right)}{\cancel{\sqrt{7}}} = 2 + 5 \sqrt{7}$