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March 12th, 2013, 07:14 AM  #1 
Newbie Joined: Feb 2013 Posts: 20 Thanks: 0  Rearranging equations  is this quite correct?
Hi guys just wondered if my working is quite correct? I needed to rearrange the equation 4t = 3k/7t 5k and make k the subject. Here's what I did; 4t = 3k/7t 5k <multiply both sides by 7t = 28t^2 = 3k  5k <add 5k to both sides = 28t^2 +5k = 3k <swap sides = 3k = 28t^2 +5k <divide both sides by three = 3k/3 = 28t^2 +5k/3 and I end up with k = 28t^2 +5k/3 Any good? 
March 12th, 2013, 07:49 AM  #2 
Senior Member Joined: Feb 2013 Posts: 281 Thanks: 0  Re: Rearranging equations  is this quite correct?
Check the first step. Are you sure you multiplied all member by 7t? Later, in one side there is 3k, in the other side there is 5k. Why don't you subtract 3k from both side? 
March 12th, 2013, 07:59 AM  #3 
Newbie Joined: Feb 2013 Posts: 20 Thanks: 0  Re: Rearranging equations  is this quite correct?
Should I have multiplied the 5k by 7t also? That does not make sense? If I subtract 3k from both sides then I end up with k still on the wrong side of the equation?

March 12th, 2013, 08:10 AM  #4 
Global Moderator Joined: Oct 2008 From: London, Ontario, Canada  The Forest City Posts: 7,766 Thanks: 1014 Math Focus: Elementary mathematics and beyond  Re: Rearranging equations  is this quite correct?
If the original equation is 4t = (3k  5k)/7t 28tē = 2k k = 14tē. If the original equation is 4t = 3k/7t  5k 28tē = 3k  35kt 28tē = k(3  35t) k = 28tē/(3  35t) 
March 12th, 2013, 08:40 AM  #5 
Newbie Joined: Feb 2013 Posts: 20 Thanks: 0  Re: Rearranging equations  is this quite correct?
3k The original equation is 4t =  5k 7t So 4t equals 3k over 7t times k6 if that makes it any more clear? I decided not to swap sides straight away, though I think I could have, My aim was to get rid of the fraction first, so I multiplied 4t by 7t on the left, to get 28t^2 then the fraction on the right by 7t to cancel it. I did not multiply 5k by 7t, I did not think I had to. I was then left with 28^t2 = 3k5k. So I then thought I should add 5k to both sides to undo the 5k, then I swapped sides and was left with 3k = 28t^2 +5k. Divided both sides by three and was left with three on its own. It made sense to me? Did I mess up somehow? Should I have multiplied the term 5k by 7t also? 
March 12th, 2013, 08:51 AM  #6 
Global Moderator Joined: Dec 2006 Posts: 18,715 Thanks: 1532 
Your equation doesn't contain k6 (or any 6). You do need to multiply 5k by 7t. After going wrong, you said you divided both sides by three and were left with three on its own. That doesn't make sense, as you ended up with k = 28t^2 +5k/3 and at no stage had three on its own. 
March 12th, 2013, 09:48 AM  #7 
Math Team Joined: Oct 2011 From: Ottawa Ontario, Canada Posts: 11,856 Thanks: 768  Re: Rearranging equations  is this quite correct?
Alex, argue all you want, BUT: if a = b/c  d then a + d = b/c so ac + cd = b 

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