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chungerforever
Group Title
Please help! How would I solve 12 5/8 + 9 2/3?? Thank you!
 one year ago
 one year ago
chungerforever Group Title
Please help! How would I solve 12 5/8 + 9 2/3?? Thank you!
 one year ago
 one year ago

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theEric Group TitleBest ResponseYou've already chosen the best response.1
Hi! Well, adding 12 and 9 are easy. Adding those fractions is a little trickier. Would you agree that \[12\frac{5}{8} + 9\frac{2}{3}\] is the same as\[12+\frac{5}{8} +9+\frac{2}{3}\]?
 one year ago

chungerforever Group TitleBest ResponseYou've already chosen the best response.0
yes, I would
 one year ago

theEric Group TitleBest ResponseYou've already chosen the best response.1
Alright, and it is also the same as \[12+9+\frac{5}{8}+\frac{2}{3}\]right?
 one year ago

theEric Group TitleBest ResponseYou've already chosen the best response.1
So 12+9, piece of cake. I can count it on my fingers (and I need some friend's fingers too). Anyway... Both fractions have to have the same bottom part, called the "denominator" if you care to remember.\[\frac{numerator}{denominator}\]
 one year ago

theEric Group TitleBest ResponseYou've already chosen the best response.1
Lets look at JUST \[\frac{5}{8}+\frac{2}{3}\]
 one year ago

chungerforever Group TitleBest ResponseYou've already chosen the best response.0
alright! (:
 one year ago

theEric Group TitleBest ResponseYou've already chosen the best response.1
Now, fiveeighths can't just be added to twothirds. You have some fractions of 8 and some fractions of 3. And since an eighth is different from a third, you can just add the top numbers.
 one year ago

theEric Group TitleBest ResponseYou've already chosen the best response.1
You have to have two fractions with the same bottom number. So they're both fractions of 24, or whatever. If you have some pieces of 24 and some other pieces of 24, you can easily know how many pieces (of 24) you have!
 one year ago

theEric Group TitleBest ResponseYou've already chosen the best response.1
So make the bottom numbers the same with simple algebra!
 one year ago

chungerforever Group TitleBest ResponseYou've already chosen the best response.0
so 15/24 + 16/24?
 one year ago

theEric Group TitleBest ResponseYou've already chosen the best response.1
Say we did want to make \[\frac{5}{8}\] a fraction of 24. Here's what we'd do: Multiply 5/8 by 1. It's the only way were not actually changing it. But that doesn't look helpful. But we can't change it. Here's the trick:\[1=\frac{1}{1}=\frac{2}{2}=\frac{3}{3}=\frac{9999}{9999}\]
 one year ago

theEric Group TitleBest ResponseYou've already chosen the best response.1
\[\frac{5}{8}*1=\frac{5}{8}*\frac{3}{3}=\frac{5*3}{8*3}=\frac{15}{24}\]
 one year ago

theEric Group TitleBest ResponseYou've already chosen the best response.1
You see how that helps? \[\frac{2}{3}*1=\frac{2}{3}*\frac{8}{8}=\frac{2*8}{3*8}=\frac{16}{24}\]
 one year ago

theEric Group TitleBest ResponseYou've already chosen the best response.1
Now its easier too add.
 one year ago

chungerforever Group TitleBest ResponseYou've already chosen the best response.0
So 21 31/24?
 one year ago

theEric Group TitleBest ResponseYou've already chosen the best response.1
Yep, but 31/24 doesn't look good. What else can it be?
 one year ago

chungerforever Group TitleBest ResponseYou've already chosen the best response.0
reduce it?
 one year ago

theEric Group TitleBest ResponseYou've already chosen the best response.1
Yeah! Wll, if by reduce it you mean turn it into a compound number and then make the numbers whole but as close to 0 as possible.. I don't remember big words like "reduce" :P Sorry!
 one year ago

theEric Group TitleBest ResponseYou've already chosen the best response.1
Go ahead and show me what you get, if you want. Or we can work it out together.
 one year ago

chungerforever Group TitleBest ResponseYou've already chosen the best response.0
Yeah I have no idea what a compound number is...
 one year ago

theEric Group TitleBest ResponseYou've already chosen the best response.1
Oh! That's one I remember! Repeat usage, I guess. It's just a number that has a whole number and a fraction. I think\[21\frac{31}{24}\] actually counts, but I know it's not a simplelooking as it can be!
 one year ago

theEric Group TitleBest ResponseYou've already chosen the best response.1
Another example is like\[5\frac{3}{4}\]
 one year ago

theEric Group TitleBest ResponseYou've already chosen the best response.1
31/24 is greater than 1, so it can be a whole number plus a fraction.
 one year ago

theEric Group TitleBest ResponseYou've already chosen the best response.1
The first step in the general way to make a fraction into a compound number is to see how many whole numbers you have. That's \[31\div24\] but not including the remainder (remainder is whatever is left over). The remainder will be written as a fraction. I know 24 goes into 31 only once. But a calculator will back me up: \[31\div24=1.2916666666666666667\]
 one year ago

theEric Group TitleBest ResponseYou've already chosen the best response.1
So there is 1 whole number in 31/24.
 one year ago

theEric Group TitleBest ResponseYou've already chosen the best response.1
Anyway..... After you take that "1" out of\[\frac{31}{24}\]it's \[\frac{31}{24}1=\frac{31}{24}\frac{24}{24}=\frac{3124}{24}=\frac{7}{24}\]
 one year ago

theEric Group TitleBest ResponseYou've already chosen the best response.1
\[21+\frac{31}{24}=21+1+\frac{7}{24}=22+\frac{7}{24}=22\frac{7}{24}\]
 one year ago

theEric Group TitleBest ResponseYou've already chosen the best response.1
So you started with\[\frac{5}{8}\]and\[\frac{2}{3}\] I seemed to just randomly choose "24" as the denominator for them both, and I also seemed to just randomly choose how to multiply each fraction. But really, I multiplied each fraction by the other's denominator over itself. \[\frac{5*2}{8*3}=\frac{15}{24}\]and\[\frac{2*8}{3*8}=\frac{16}{24}\] so the "24" kinda just happened.
 one year ago

theEric Group TitleBest ResponseYou've already chosen the best response.1
I hope I've helped more than I've hurt! I feel like my response has been jumpled, so I'm sorry. Ask me any questions you have, and I'll try to be more clear.
 one year ago

chungerforever Group TitleBest ResponseYou've already chosen the best response.0
thank you so much!
 one year ago

theEric Group TitleBest ResponseYou've already chosen the best response.1
You're welcome. I hope you can do the problem on your own now!
 one year ago
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