# Reducing a fraction by dividing top and bottom

I am trying to cancel out to reduce this:

$$\frac{ 6xh + 3h^2 + 5h }{ h }$$

Is it possible to cancel out the h's to become this; $$\frac{ 6x + 3 + 5 }{ h }$$

While the $3$ goes into the $6x$? So the final answer is $2x + 5$?

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No. I'm not sure I can give more of an answer without knowing why you think you would be able to do that. – Callus Mar 12 '14 at 2:34
After factoring $h$ out to get $\frac{h(6x+3h+5)}{h}$ (which is just the distributive property), you can cancel the common $h$ factor in the numerator and denominator to get $6x+3h+5$, which can't really be simplified any further. – Dustan Levenstein Mar 12 '14 at 2:35
$\frac{6xh+3h^2+5h}{h}=\frac{h(6x+3h+5)}{h}=6x+5+3h$ – Eleven-Eleven Mar 12 '14 at 2:35

You got a good idea, but it needs some cleaning up.

What you want to do is factor out the h. This will help you to cancel out the common $h$ in both the top and bottom of the problem. It would look like this,

$$\frac{ 6xh + 3h^2 + 5h }{ h }=\frac{ h(6x + 3h + 5) }{ h }$$

Assuming that $h\neq0$, we can cancel it out. Now you will get $6x+3h+5$ which can not be simplified further.

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