anonymous
  • anonymous
Please help! How would I solve 12 5/8 + 9 2/3?? Thank you!
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
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theEric
  • theEric
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\[12+\frac{5}{8} +9+\frac{2}{3}\]?
anonymous
  • anonymous
yes, I would
theEric
  • theEric
Alright, and it is also the same as \[12+9+\frac{5}{8}+\frac{2}{3}\]right?

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theEric
  • theEric
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.\[\frac{numerator}{denominator}\]
theEric
  • theEric
Lets look at JUST \[\frac{5}{8}+\frac{2}{3}\]
anonymous
  • anonymous
alright! (:
theEric
  • theEric
Now, five-eighths can't just be added to two-thirds. 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.
theEric
  • theEric
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!
theEric
  • theEric
So make the bottom numbers the same with simple algebra!
anonymous
  • anonymous
so 15/24 + 16/24?
theEric
  • theEric
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:\[1=\frac{1}{1}=\frac{2}{2}=\frac{3}{3}=\frac{9999}{9999}\]
theEric
  • theEric
\[\frac{5}{8}*1=\frac{5}{8}*\frac{3}{3}=\frac{5*3}{8*3}=\frac{15}{24}\]
theEric
  • theEric
You see how that helps? \[\frac{2}{3}*1=\frac{2}{3}*\frac{8}{8}=\frac{2*8}{3*8}=\frac{16}{24}\]
theEric
  • theEric
Now its easier too add.
anonymous
  • anonymous
So 21 31/24?
theEric
  • theEric
Yep, but 31/24 doesn't look good. What else can it be?
anonymous
  • anonymous
reduce it?
theEric
  • theEric
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!
theEric
  • theEric
Go ahead and show me what you get, if you want. Or we can work it out together.
anonymous
  • anonymous
Yeah I have no idea what a compound number is...
theEric
  • theEric
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 simple-looking as it can be!
theEric
  • theEric
Another example is like\[5\frac{3}{4}\]
theEric
  • theEric
31/24 is greater than 1, so it can be a whole number plus a fraction.
theEric
  • theEric
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: \[31\div24=1.2916666666666666667\]
theEric
  • theEric
So there is 1 whole number in 31/24.
theEric
  • theEric
Anyway..... After you take that "1" out of\[\frac{31}{24}\]it's \[\frac{31}{24}-1=\frac{31}{24}-\frac{24}{24}=\frac{31-24}{24}=\frac{7}{24}\]
theEric
  • theEric
\[21+\frac{31}{24}=21+1+\frac{7}{24}=22+\frac{7}{24}=22\frac{7}{24}\]
theEric
  • theEric
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.
theEric
  • theEric
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.
anonymous
  • anonymous
thank you so much!
theEric
  • theEric
You're welcome. I hope you can do the problem on your own now!

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