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August 28th, 2009, 12:16 PM  #1 
Senior Member Joined: Jul 2009 Posts: 136 Thanks: 0  simultaneous equations anyone?
OK, I got this one I believe. y/2  x = 2, 6x  3y/2 = 3 and I ended up with y = 10, x = 3. But these two are giving me fits: x/2  y/5 = 1, y  x/3 = 8. I started by simplifying the y as y = 8 + x/3 and substituting that into the other "y" that would present itself as: x/2  8/5 + x/3/5 = 1. The "x/3/5" should be read as x/3 as the numerator and 5 as the denominator. So I suppose I need to know if that is correct as a start. Let me know, thanks! IF you can go further with it, then that would be great. 
August 28th, 2009, 12:57 PM  #2 
Global Moderator Joined: Oct 2008 From: London, Ontario, Canada  The Forest City Posts: 7,540 Thanks: 920 Math Focus: Elementary mathematics and beyond  Re: simultaneous equations anyone?
I did it this way: x/2  y/5 = 1, y  x/3 = 8 ? 10(x/2  y/5) = 10, 3(y  x/3) = 24 ? 5x  2y = 10, 3y  x = 24 . . . . and so on . . . . 
August 28th, 2009, 04:34 PM  #3 
Senior Member Joined: Apr 2007 Posts: 2,140 Thanks: 0 
Since is equivalent to saying that , and is equivalent to saying that since is obvious, the equation is true, and solving the equation gives us and we will use this value to determine the obvious of 
August 30th, 2009, 04:25 AM  #4 
Senior Member Joined: Jul 2009 Posts: 136 Thanks: 0  Re: simultaneous equations anyone?
Thanks everyone! Wouldn't you know that the very next day I got it! y = 8 + x/3 as noted earlier and then I followed it up by simplifying: 5x/10  16  2x/3 /10 = 1... 5x  16  2x/3 = 10... 15x/3  16  2x/3 = 10... 15x/3  2x/3 = 26... 13x/3 = 26... 13x = 78... x = 6... I appreciate your help and assistance.... Have been enjoying the problems! 
August 30th, 2009, 04:28 AM  #5 
Senior Member Joined: Apr 2007 Posts: 2,140 Thanks: 0 
IMO, equations are fun to study! 

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