Foundations of GMAT Math (5th ed.), Ch. 17, Q. 19

[InayatC54]

Here is the text of the question (EDITED after reply #1):
Simplify the following expression:

If t is not equal to 1/2,

2t-1+ (2t-1)^2
---------------------
2t-1

(A) 2t (B) 2t-1 (C) 2t-2

This is how I solved it:
(2t-1)^2 (because the (2t-1) from both the numerator and denominator cancel out.
This is where I got stuck.
I was going to expand this as I identified it as the square of a difference special product, yielding: 4t^2+4t+1, but that is not one of the answer choices provided.

The solution in the book shows:
1+(2t-1) = 2t
I'm confused about where the 1 comes from. Basically, how do you get from: (2t-1)^2 to 1+(2t-1)

[tim]

Pay closer attention to what the book actually says (note that what you have written is not accurate). If you can transcribe the problem properly here on the forum, you may see your error. If not, we will be glad to help you further once you have written the problem like it shows up in the book. My advice would be either to use more parentheses or make a huge fraction bar like this:

12345
----------
67890

[RonPurewal]

you know that (1 + x)/x is not equal to 1, right?

if you know that, then you should be able to see your error here.
if a fraction contains a sum/difference, then you can only 'cancel' things you can factor out of EVERY term of the sum/difference.

[InayatC54]

Geez! I see it now. Thanks.

[tim]

Glad to hear it! Let us know if you have any further questions on this one.

[RonPurewal]

Geez! I see it now. Thanks.

this, by the way, is the key to understanding anything more complex: first, investigate simpler versions of the same idea.

you'll find that, as you move up the scale, the concepts themselves get more difficult only rarely. more likely, you'll just keep doing the same things with more stuff.