My Math Forum Fractions issue

 Linear Algebra Linear Algebra Math Forum

 February 4th, 2017, 02:28 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, 02:39 PM #2 Global Moderator   Joined: May 2007 Posts: 6,394 Thanks: 545 4n=4n-3+3. The rest is easy. Thanks from Country Boy
 February 4th, 2017, 02:41 PM #3 Newbie   Joined: Feb 2017 From: UK Posts: 2 Thanks: 0 Cheers mate
 February 4th, 2017, 11:55 PM #4 Senior Member     Joined: Apr 2014 From: Europa Posts: 571 Thanks: 175 $\displaystyle n\in \mathbb{N} \\\;\\ \dfrac{4n}{4n-3}=\dfrac{4n-3+3}{4n-3}=\dfrac{4n-3}{4n-3}+\dfrac{3}{4n-3} = 1 +\dfrac{3}{4n-3}$

 Tags fractions, issue

 Thread Tools Display Modes Linear Mode

 Similar Threads Thread Thread Starter Forum Replies Last Post Frontier111 Physics 9 May 13th, 2016 04:18 AM Chickwolf Linear Algebra 9 November 12th, 2015 06:03 PM bigcity Algebra 3 January 2nd, 2012 12:09 PM nova3421 Algebra 1 September 9th, 2009 11:03 AM baxy7 Applied Math 1 June 1st, 2009 06:43 AM

 Contact - Home - Forums - Cryptocurrency Forum - Top