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February 4th, 2017, 02:28 PM   #1
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Fractions issue

Hi,

I am sure I am missing a trick here, but could someone please help me understand why,

$\displaystyle \frac{4n}{4n-3}=1+\frac{3}{4n-3}$

Thank you in advance,

Matt
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February 4th, 2017, 02:39 PM   #2
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4n=4n-3+3. The rest is easy.
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February 4th, 2017, 02:41 PM   #3
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Cheers mate
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February 4th, 2017, 11:55 PM   #4
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$\displaystyle n\in \mathbb{N}
\\\;\\
\dfrac{4n}{4n-3}=\dfrac{4n-3+3}{4n-3}=\dfrac{4n-3}{4n-3}+\dfrac{3}{4n-3} = 1 +\dfrac{3}{4n-3}$
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