1. SAT fractions

Hi, I was doing some problems in an SAT math workbook and came across this:
(all parentheses added by me to deter confusion)
1/(1+1/b) = p Which expression is 2p?
The answer given by the book is 1/ ((1/2)+(1/2b))

However, it would seem to me if p=1/(1+1/b) then 2p=2(1/(1+1/b)) = 2/(2(1+1/b)) = 2/(2+2/b)

Can someone explain this to me please?

Hi, I was doing some problems in an SAT math workbook and came across this:
(all parentheses added by me to deter confusion)
1/(1+1/b) = p Which expression is 2p?
The answer given by the book is 1/ ((1/2)+(1/2b))

However, it would seem to me if p=1/(1+1/b) then 2p=2(1/(1+1/b)) = 2/(2(1+1/b)) = 2/(2+2/b)

Can someone explain this to me please?
$\displaystyle 2p = \frac{p}{\frac{1}{2}} = \frac{1}{\frac{1}{2}} \cdot p = \frac{1}{\frac{1}{2}} \cdot \frac{1}{1 + \frac{1}{b}} = \frac{1}{\frac{1}{2} + \frac{1}{2b}}$

3. Thanks Skeeter for the reply. I understand now how the book got the answer 1/ ((1/2)+(1/2b)). But I still don't understand logically why if p = something that 2*p = 2*something doesn't work. I know I'm probably just missing something incredibly stupid, but I just don't get it.

$\displaystyle 2 \cdot \frac{1}{1 + \frac{1}{b}} = \frac{2}{1 + \frac{1}{b}}$ , not $\displaystyle \frac{2}{2 + \frac{2}{b}}$
$\displaystyle \frac{2}{2 + \frac{2}{b}} = p$