math problem?
thanks so much for helping guys. im starting to understand this all a little better.
heres another one.. and can you please show work so that i can actually understand how to get the answer? then maybe i wont have to keep asking people on here. haha.
remember.. best answer points!
heres a picture of the problem:
http://i11.tinypic.com/6y4yvpl.jpg
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ok, so we have a fraction being added to another fraction. the only difference is that we have varialbes. but that doesn't change anything.
So, 1st let's find the common denominator.
the least common multiple between 8 and 10 is 80. So we must multiply the first fraction by (10/10) and the 2nd fraction by (8/8)
doing so, we get:
(50b+20)/80 + (24b-32)/80
combine terms:
(74b -12)/80 divide all terms by 2 to get the answer:
(37b-6)/40 which is option 1
(5b + 2) / 8 + (3b - 4)/10 =
The common denominator is 40
(25b + 10) / 40 + (12b - 16)/40 =
(37b - 6 )/ 40
- - - - - - -
The answer is option 1
- - - - - - - -s-
(5b+2)/8 + (3b - 4)/10 =
10/10 * (5b+2)/8 + 8/8 * (3b - 4)/10 =
10 * (5b+2)/(8*10) + 8 * (3b - 4)/ (10 * 8) =
(50b + 20 + 24 b - 32)/80 =
(74b - 12)/80 =
2(37b - 6) / (2*40) =
(37b - 6)/40
(5b + 2) / 8 + (3b-4) / 10
= 5(5b+2) / 40 + 4(3b-4) /40
= [ 5(5b+2) + 4(3b-4) ] /40
= (37b - 6) /40
Yor correct answer is --Option 1
whoa......... how do u do that stuff? i'm only in fifth grade.... i'm sick today.