Guide 6, Page 48, on the bottom

[Pnwankpa]

This particular example has been a point of confusion for me for some time on page 48 on the bottom. I have tried to figure it out from time to time, and I can never wrap my head around how they arrived at 4v - 2 - 1, as you solve for the problem. In this problem, we are given examples of strange formulas that could pop up on the GRE:

v& = 2v - 1

Quantity A Quantity B
(v&)& 4v

Using the formula given, I plugged it into Quantity A, and got the following:
(v&)& => (2v-1)& => 2(2v-1) - 1& => 4v - 2 - &;

However, in the book, they arrived at the following:
(2v - 1)& = 2(2v - 1) - 1,
I went back and evaluated the original expression, and realized that if I solved for '&,' I get that it equals one. With that in mind, I attempted to solve again by plugging in the known variables, but what I arrived at was: (v&)& = (2v-1)(1) => 2v -1; much different from 4v-3, as was shown in the book. So my question is, how do I know when to plug in the value of a variable if known? In the book, even though it was clear that v& = 2v - 1, and & = 1, they substituted for 'v&', yet left '&' until only after they had distributed (2v-1)& => 2v& - &. I'm sure that there is a rule about this that I just do not remember. Your clarification on this will be greatly appreciated.

Thank you for your time.

[tommywallach]

I've actually answered this question a couple times in the forum already. If you look through, you'll find it. Let me know if you have a problem locating it.

-t