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!!
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