Canceling a Common Factor

[Drakkith]

Homework Statement

I've got a fraction here:

$\frac{14+14\sqrt{3}}{-8}$

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

$\frac{7+7\sqrt{3}}{-4}$

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).

 

[Mentallic]

What does
$$\frac{14y}{2x}$$
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.

 

[SteamKing]

Really?

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

 

[Drakkith]

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.

 

[SteamKing]

Multiplication distributes over addition, so a*(b + d) = a*b + a*d

 

[Mentallic]

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

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 $14(1+\sqrt{3})$ so we can now let $y=1+\sqrt{3}$.

 

[Drakkith]

So in this case, the numerator is $14(1+\sqrt{3})$ so we can now let $y=1+\sqrt{3}$.

Arrghh... I had my answer as $\frac{7(1+\sqrt{3})}{4}$ , which was apparently wrong, whereas $\frac{7+7\sqrt{3})}{4}$ was correct.

 

[Office_Shredder]

Arrghh... I had my answer as $\frac{7(1+\sqrt{3})}{4}$ , which was apparently wrong, whereas $\frac{7+7\sqrt{3})}{4}$ was correct.

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

 

[Drakkith]

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...