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 February 4th, 2017, 01:28 PM #1 Newbie   Joined: Feb 2017 From: UK Posts: 2 Thanks: 0 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
 February 4th, 2017, 01:39 PM #2 Global Moderator   Joined: May 2007 Posts: 6,524 Thanks: 587 4n=4n-3+3. The rest is easy. Thanks from Country Boy
 February 4th, 2017, 01:41 PM #3 Newbie   Joined: Feb 2017 From: UK Posts: 2 Thanks: 0 Cheers mate
 February 4th, 2017, 10:55 PM #4 Senior Member     Joined: Apr 2014 From: Europa Posts: 571 Thanks: 176 $\displaystyle n\in \mathbb{N} \\\;\\ \dfrac{4n}{4n-3}=\dfrac{4n-3+3}{4n-3}=\dfrac{4n-3}{4n-3}+\dfrac{3}{4n-3} = 1 +\dfrac{3}{4n-3}$

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