# A simple simplification problem

• Dec 6th 2006, 10:34 PM
shenton
A simple simplification problem
I thought I know this one but the answer stumps me:

(3a - 3b) / (6a - 6b)

After cancelling out the "a" and "b", I would think I have (3 - 3) in the numerator and (6 - 6) in the denominator and the answer would be 0. How did the answer turns out to 1/2.

I think I'm missing some concept here. Pls help. Thanks.
• Dec 7th 2006, 12:12 AM
AfterShock
Quote:

Originally Posted by shenton
I thought I know this one but the answer stumps me:

(3a - 3b) / (6a - 6b)

After cancelling out the "a" and "b", I would think I have (3 - 3) in the numerator and (6 - 6) in the denominator and the answer would be 0. How did the answer turns out to 1/2.

I think I'm missing some concept here. Pls help. Thanks.

(3*a - 3*b)/(6*a - 6*b) = 1/2

I can see why you think it is 0. You are thinking the problem is:

(3*a)/(6*b) - (3*a)/(6*b) = 0.

However, this problem is:

(3*a)/(6*a - 6*b) - (3*b)/(6*a - 6*b)

Do you see they each have a common denominator. Now you can cancel to get:

(3*a)/(6*a - 6*b) = a/[2(a - b)] and:

(3*b)/(6*a - 6*b) = b/[2(a - b)]

Thus, a/[2(a - b)] - b/[2(a - b)] = (a - b)/[2(a - b)]

Cancel the (a - b) from the numerator and denominator. Note that there is a "1 times (a - b)" at the top; thus, you are left with 1/2.

Do you see why it isn't 0?
• Dec 7th 2006, 08:43 AM
shenton
Thanks for the detailed explanation.

Things certainly does not look as simple as it seems.
• Dec 7th 2006, 11:24 AM
topsquark
Quote:

Originally Posted by shenton
I thought I know this one but the answer stumps me:

(3a - 3b) / (6a - 6b)

After cancelling out the "a" and "b", I would think I have (3 - 3) in the numerator and (6 - 6) in the denominator and the answer would be 0. How did the answer turns out to 1/2.

I think I'm missing some concept here. Pls help. Thanks.

A simpler explanation is this (but it involves factoring):
$\displaystyle \frac{3a - 3b}{6a - 6b} = \frac{3(a-b)}{6(a-b)} = \frac{3}{6} = \frac{1}{2}$ (Requires $\displaystyle b \neq a$!)

-Dan
• Dec 7th 2006, 11:29 AM
AfterShock
Why can't we have a = b? We can use that guys new nullity number and divide by 0! :p
• Dec 7th 2006, 11:31 AM
topsquark
Quote:

Originally Posted by AfterShock
Why can't we have a = b? We can use that guys new nullity number and divide by 0! :p

:eek: (groans in great agony)

-Dan