How to Simplify Fractions with Common Factors

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Homework Statement



I've got a fraction here:

[itex]\frac{14+14\sqrt{3}}{-8}[/itex]

Why is it you can take a 2 out of the bottom and top to make it the following?

[itex]\frac{7+7\sqrt{3}}{-4}[/itex]

I'm lost in figuring out how this works. I thought the top was like having (14+14x), where you can take a 14 out of each term and make it 14(1+x).
 
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What does
[tex]\frac{14y}{2x}[/tex]
equal to?

Also, you can move the negative sign from the denominator to the numerator by "taking out" a -1 from both the numerator and denominator.
 
Really?

Code:
Lookit:
14 + 14*Sqrt(3)      2*7 + 2*7 * Sqrt(3)     2*[7 + 7 * Sqrt (3)]     7 + 7 * Sqrt (3)
---------------- =   ------------------- =   -------------------- = ----------------
     -8                     2*(-4)                   2 * (-4)               -4
 
I understand that 14y/2x = 7y/x.

I thought you couldn't divide the original fraction that way because it's still adding up there and you had to take a factor out or something first.
 
Multiplication distributes over addition, so a*(b + d) = a*b + a*d
 
Drakkith said:
I understand that 14y/2x = 7y/x.

I thought you couldn't divide the original fraction that way because it's still adding up there and you had to take a factor out or something first.

Yes, that's true, but look at what you said earlier

Drakkith said:
I'm lost in figuring out how this works. I thought the top was like having (14+14x), where you can take a 14 out of each term and make it 14(1+x).

So in this case, the numerator is [itex]14(1+\sqrt{3})[/itex] so we can now let [itex]y=1+\sqrt{3}[/itex].
 
Mentallic said:
So in this case, the numerator is [itex]14(1+\sqrt{3})[/itex] so we can now let [itex]y=1+\sqrt{3}[/itex].

Arrghh... I had my answer as [itex]\frac{7(1+\sqrt{3})}{4}[/itex] , which was apparently wrong, whereas [itex]\frac{7+7\sqrt{3})}{4}[/itex] was correct.
 
Drakkith said:
Arrghh... I had my answer as [itex]\frac{7(1+\sqrt{3})}{4}[/itex] , which was apparently wrong, whereas [itex]\frac{7+7\sqrt{3})}{4}[/itex] was correct.

Those are both the exact same number, and are equally correct actually (well, except for the missing minus sign)
 
Office_Shredder said:
Those are both the exact same number, and are equally correct actually (well, except for the missing minus sign)

Ah yes, forgot the negative.
At least I got it figured out. I was so confused...