• Oct 28th 2010, 09:49 PM
piercedgeek
I'm a little embarrassed, this question looks so simple, but I'm obviously forgetting something basic when trying to solve it, because I end up not making it to the final answer in the book.

Question:
$ab + a = 1$ (for a)

$ab = 1 - a$ (minus a from both sides)

$a = (1-a)/b$ (divide by b)

I know this isn't the final answer, because we still have a on both sides, but I just feel stuck. I know I can make it 1/b - a/b, but can't remember how to get rid of the a. Did I forget something simple here? or do something incorrect earlier that has painted me in to an impossible corner?

$a = 1/(b+1)$
• Oct 28th 2010, 10:24 PM
Wilmer
Quote:

Originally Posted by piercedgeek
$ab + a = 1$ (for a)

a(b + 1) = 1
a = 1 / (b + 1)
• Oct 28th 2010, 10:27 PM
piercedgeek
ok, after reviewing some old examples, I see I can use the distributive property
to re-write ab+a as a(1+b), so,
$a(1+b) = 1$ (divide by 1+b)

$a = 1/(1+b)$

and it's done...

But I think this is the source of a lot of the problems I have with questions that have multiple variables.
The book would describe this as "using the distributive property to write the sum of ab+a as a product of a and 1+b"
But for some reason I just have a hard time seeing this particular method in my head, so I freeze when a problem requires it... Does anyone have another way to view this?
• Oct 28th 2010, 10:40 PM
piercedgeek
Sorry to be almost talking to myself on this forum, I've been staring at this problem for longer than I care to admit, with no progress, and after talking/typing things out, the gears are starting to turn in my head. It just hit me, and I figured I'd share in case anyone else is having a brain fart moment like myself:
a+ab can be written as a(1+b) because a(1+b) can be distributed to a+ab

could be time for bed