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chungerforever
Please help! How would I solve 12 5/8 + 9 2/3?? Thank you!
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\[12+\frac{5}{8} +9+\frac{2}{3}\]?
Alright, and it is also the same as \[12+9+\frac{5}{8}+\frac{2}{3}\]right?
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.\[\frac{numerator}{denominator}\]
Lets look at JUST \[\frac{5}{8}+\frac{2}{3}\]
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.
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!
So make the bottom numbers the same with simple algebra!
so 15/24 + 16/24?
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:\[1=\frac{1}{1}=\frac{2}{2}=\frac{3}{3}=\frac{9999}{9999}\]
\[\frac{5}{8}*1=\frac{5}{8}*\frac{3}{3}=\frac{5*3}{8*3}=\frac{15}{24}\]
You see how that helps? \[\frac{2}{3}*1=\frac{2}{3}*\frac{8}{8}=\frac{2*8}{3*8}=\frac{16}{24}\]
Now its easier too add.
Yep, but 31/24 doesn't look good. What else can it be?
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!
Go ahead and show me what you get, if you want. Or we can work it out together.
Yeah I have no idea what a compound number is...
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!
Another example is like\[5\frac{3}{4}\]
31/24 is greater than 1, so it can be a whole number plus a fraction.
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: \[31\div24=1.2916666666666666667\]
So there is 1 whole number in 31/24.
Anyway..... After you take that "1" out of\[\frac{31}{24}\]it's \[\frac{31}{24}-1=\frac{31}{24}-\frac{24}{24}=\frac{31-24}{24}=\frac{7}{24}\]
\[21+\frac{31}{24}=21+1+\frac{7}{24}=22+\frac{7}{24}=22\frac{7}{24}\]
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.
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.
thank you so much!
You're welcome. I hope you can do the problem on your own now!