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September 22nd, 2016, 03:26 PM  #1 
Senior Member Joined: Feb 2016 From: seattle Posts: 377 Thanks: 10  promise last one for today, 3+x/3 >= x/51
3+x/3 >= x/51 so I thought I would get 5x/15 3x/15 =2x/15 >= 2 Yet besides that I am getting them all wrong, I don't think this is right because how would I get rid of the 2x/15? 
September 22nd, 2016, 03:43 PM  #2  
Math Team Joined: Jul 2011 From: Texas Posts: 2,780 Thanks: 1431  Quote:
$45 + 5x \ge 3x  15$ finish ...  
September 23rd, 2016, 03:02 PM  #3 
Global Moderator Joined: May 2007 Posts: 6,641 Thanks: 625  $\displaystyle \frac{2x}{15}\ge 2\Rightarrow 2x\ge 30\Rightarrow x \ge 15$. I presume the problem was to get an inequality in x.

October 24th, 2016, 02:12 AM  #4 
Member Joined: Sep 2016 From: India Posts: 88 Thanks: 30 
$3+\dfrac{x}{3}\geq\dfrac{x}{5}1$ Multiply each side by $15$ $\implies45+5x\geq3x15$ Separate the like terms, $\implies 5x3x\geq4515$ $\implies 2x\geq30$ $\implies x\geq15$ Last edited by deesuwalka; October 24th, 2016 at 02:18 AM. 

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