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I have a fraction I am trying to solve. I know the answer already, as Wolfram says it is $\frac{143}{300}$.
The fraction is: $$\frac{5}{12} + \frac{3}{50} = \space ?$$ Please explain why and how your method works.

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Null has given you a good way. Here's a way without worrying about the LCM: $${a\over b}+{c\over d}={ad+bc\over bd}$$ In the example, $${5\over12}+{3\over50}={(5)(50)+(12)(3)\over(12)(50)}={286\over600}={143\over300}$$

The price of not worrying about the LCM is that you get an answer, $286/600$, that isn't in lowest terms, so you have the extra step at the end of reducing the fraction.

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  • $\begingroup$ I've tried that method, it doesn't always work? $\endgroup$ – kinesis Oct 17 '14 at 6:30
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    $\begingroup$ It does always work [in fact it is usually taken as the definition of addition], but sometimes it doesn't give you the answer in simplest form (lowest terms). $\endgroup$ – Eric Stucky Oct 17 '14 at 6:45
  • $\begingroup$ Then again, no method is guaranteed to give the answer in lowest terms. E.g., $(1/2)+(1/2)=2/2=1$. $\endgroup$ – Gerry Myerson Oct 17 '14 at 10:57
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The fractions need to have the same denominator in order to add them together. The denominator must be the least common multiple of the two denominators, or a multiple of it. The least common multiple of 12 and 50 is 300.

First multiply $5/12$ by $25/25$. Since $25/25 = 1$ the product is the same as the original fraction. You simply multiply the numerators together and the denominators together:

$$\frac{5}{12}\times\frac{25}{25} = \frac{125}{300}$$

Then multiply $3/50$ by $6/6$. Again, the product is the same as the original fraction:

$$\frac{3}{50}\times\frac{6}{6} = \frac{18}{300}$$

Now both fractions have the same denominator and you can simply add the numerators:

$$\frac{125}{300}+\frac{18}{300} = \frac{143}{300}$$

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  • $\begingroup$ What's the significance of the 25? Where do you pull this from? Why did you divide 50 by 2? $\endgroup$ – kinesis Oct 17 '14 at 6:10
  • $\begingroup$ @kinesis You need the denominators to be equal to do the addition. The multiples of 50 (the other denominator) are 100, 150, 200, etc. The first multiple of 12 (the first denominator) that is equal to a multiple of 50 is 300, which requires multiplying by 25. Ultimately, I'm multiplying the denominators until they equal the least common multiple, which in this case is 300. $\endgroup$ – Null Oct 17 '14 at 6:13
  • $\begingroup$ Great. When would you not use LCM to get equal denominators? Isn't it ineffective to have to write down all the multiples on a test? $\endgroup$ – kinesis Oct 17 '14 at 6:18
  • $\begingroup$ @kinesis Generally you use the LCM but you can use any multiple of it (maybe you come up with a multiple of it first, or maybe it's easier to work with). For example, I could have simply multiplied 12 by 100 to get 1200, which would require multiplication of 50 by 24 to get 1200 for it as well. It doesn't matter what you use as long as you use the same denominator for the two fractions. $\endgroup$ – Null Oct 17 '14 at 6:21
  • $\begingroup$ Isn't it ineffective to have to write down all the multiples while doing a test? I see.. so any multiple ... $\endgroup$ – kinesis Oct 17 '14 at 6:21
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Here is another equivalent idea, put $$x=\frac{5}{12}+\frac{3}{50}$$ Multiply by 12 $$12x=5+\frac{36}{50}=5+\frac{18}{25}$$ Multiply by 25 $$300x=125+18$$ Divide by 300 $$x=\frac{143}{300}$$

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If $a=b$ then for any function $f(a)=f(b)$. Suppose $\displaystyle x=\frac{5}{12}+\frac{3}{50}$. Then $\displaystyle (12\cdot 50)\cdot x=(12\cdot 50)\cdot\left(\frac{5}{12}+\frac{3}{50}\right)$, so $\displaystyle 600x=\frac{12\cdot 50\cdot 5}{12}+\frac{12\cdot 50\cdot 3 }{50}=$ $=50\cdot 5+12\cdot 3=286$, why $\displaystyle x=\frac{286}{600}=\frac{143}{300}$.

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