PDA

View Full Version : For all you math whizzes, I need help with a Singapore problem . . .


Jackie in AR
02-29-2008, 12:51 PM
[2(a-b)] - [(a/3 + b/4) + (a/6 - 3b/4)] =

The answer is 3/2(a-b) according to the book.

Will someone please show me how to work this problem correctly?

Thanks!

Krista in LA
02-29-2008, 01:10 PM
[2(a-b)] - [(a/3 + b/4) + (a/6 - 3b/4)] =

The answer is 3/2(a-b) according to the book.

Will someone please show me how to work this problem correctly?

Thanks!

The lowest common demonitator for 3, 4, and 6 is 12 so you need to multiply each fraction to get it over 12.

(12/12)* [2(a-b)] - {[(4/4)(a/3) + (3/3)(b/4)] + [(2/2)(a/6) - (3/3)(3b/4)]} =

24(a-b)/12 - [(4a/12) + (3b/12) + 2a/12 - 9b/12)] =
(24a-24b)/12 - [(6a-6b)/12] =
(18a-18b)/12 =
18(a-b)/12 =
3(a-b)/2 = 3/2 (a-b)

Snickerdoodle
02-29-2008, 01:13 PM
:)

Kat in GA
02-29-2008, 01:18 PM
[2(a-b)] - [(a/3 + b/4) + (a/6 - 3b/4)] =

The answer is 3/2(a-b) according to the book.

Will someone please show me how to work this problem correctly?

Thanks!

Here is a slight variation on the other answer:

First get rid of the brackets:
2a - 2b - a/3 - b/4 - a/6 + 3b/4

Change all the a's to sixths and all the b's to fourths:

12a/6 - 8b/4 - 2a/6 - b/4 - a/6 + 3b/4 = 9a/6 - 6b/4

Then you reduce: 3a/2 - 3b/2 = 3/2 (a - b)

JenneinAZ
02-29-2008, 01:27 PM
[2(a-b)] - [(a/3 + b/4) + (a/6 - 3b/4)] =

distributing the 2 and the negative: 2a-2b -a/3 -b/4 -a/6 +3b/4 =

Common denominator of 12: 24a/12 -24b/12 -4a/12 -3b/12 -2a/12 +9b/12

Reorder to put like terms together: 24a/12 -4a/12 -2a/12 -24b/12 -3b/12 +9b/12

Combine like terms: (24a-4a-2a)/12 + (-24b-3b+9b)/12

More combining: 18a/12 + (-18b)/12

Simplifying: 3a/2 -3b/2

More simplifying: (3a-3b)/2

Still more: (3/2)(a-b)

Jackie in AR
02-29-2008, 01:41 PM
Here is what I did wrong: I multiplied each term by 12 instead of 12/12 as I should have.

And when I divide the answer I came up with (18a-18b) by 12, I get 3/2 (a-b).

Many thanks to all of you!!