My Math Forum stuck on problem

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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)x-5 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: 19,162 Thanks: 1638 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
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Quote:
 Originally Posted by adriana.alrb 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.
We know that x + y = 8 + (-1) = 7. Is the second sentence mathematically stated as (x + y)(xy)^4, or {xy(x + y)}^4?

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