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August 30th, 2009, 08:35 AM  #1 
Newbie Joined: Aug 2009 Posts: 8 Thanks: 0  stuck on problem
Please help me solve this problem. I've been working on this problem,and i just keep reducing it to Y=(3/2)x5 and x=(2/3)y+(10/3) and i don't know what to do next. i haven't gotten to the second part because i am stuck with the first part. I'd be grateful if someone could help me (problem below). Solve this system for x and y. Find the sum of your x and y values. Multiply your answer by the product of your x and y values raised to the fourth power. 
August 30th, 2009, 09:45 AM  #2 
Member Joined: Jul 2009 Posts: 34 Thanks: 0  Re: stuck on problem
First multiply both equations by 6. Then solve for x in the second equation, which will result in an expression in y. Replace x in the first equation by this expression, and it is easy to solve for y. After that it is also easy to find x. 
August 30th, 2009, 01:50 PM  #3 
Newbie Joined: Aug 2009 Posts: 8 Thanks: 0  Re: stuck on problem
i still don't get it i need more help. PLEASE!!! 
August 30th, 2009, 02:39 PM  #4 
Global Moderator Joined: Dec 2006 Posts: 18,048 Thanks: 1395 
You need to be more careful. From the first equation, x/2 + 1  (y/3 + 4/3) = 4. Hence x/2 = y/3 + 4/3 + 4  1 = y/3 + 13/3. From the second equation, x/2 + y/2 = 1/2 + x/3  y/3. Hence x/2  x/3 = 1/2  y/2  y/3, and so x/2 = 3/2  5y/2. It follows that y/3 + 13/3 = 3/2  5y/2, and so y/3 + 5y/2 = 3/2  13/3, i.e., 17y = 17, so y = 1. Hence x/2 = y/3 + 13/3 = 1/3 + 13/3 = 4, so x = 8. 
August 30th, 2009, 02:53 PM  #5 
Newbie Joined: Aug 2009 Posts: 8 Thanks: 0  Re: stuck on problem
THANK YOU SOOOOOOO MUCH!! YOU my friend are a GENIUS!!!!! I cant thank you guys enough for helping me out THANKS AGAIN 
August 30th, 2009, 10:14 PM  #6  
Senior Member Joined: Apr 2007 Posts: 2,140 Thanks: 0  Quote:
 

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