August 24th, 2015, 07:29 PM  #1 
Newbie Joined: Aug 2015 From: vernal Posts: 1 Thanks: 0  Math Homework
can someone help me do this problem for my sons homework. 1/3(2x1) = 1/2(3x+1)

August 24th, 2015, 08:19 PM  #2 
Global Moderator Joined: Oct 2008 From: London, Ontario, Canada  The Forest City Posts: 7,697 Thanks: 977 Math Focus: Elementary mathematics and beyond 
1/3(2x  1) = 1/2(3x + 1) Multiply both sides by the least common multiple of 2 and 3, which is 6: 2(2x  1) = 3(3x + 1) Expand: 4x  2 = 9x + 3 Subtract 4x from each side: 2 = 5x + 3 Subtract 3 from each side: 5 = 5x Solve for x: x = 5/5 = 1 
August 25th, 2015, 06:02 AM  #3 
Math Team Joined: Jan 2015 From: Alabama Posts: 2,922 Thanks: 785 
Did you mean $\displaystyle \frac{1}{3}(2x 1)= \frac{1}{2}(3x 1)$, which is what greg1313 assumed, or did you mean $\displaystyle \frac{1}{3(2x 1)}= \frac{1}{2(3x 1)}$, which is what I might have been inclined to think. For both you start by "getting rid of the denominators". In the first you multiply both sides by (3)(2)= 6. For the second, multiply by 6(2x 1)(3x 1). 
August 25th, 2015, 06:06 AM  #4 
Global Moderator Joined: Dec 2006 Posts: 18,444 Thanks: 1462 
$\displaystyle \frac{1}{3(2x 1)} = \frac{1}{2(3x 1)}$ has no solutions.

August 25th, 2015, 06:21 AM  #5 
Math Team Joined: Dec 2013 From: Colombia Posts: 7,118 Thanks: 2369 Math Focus: Mainly analysis and algebra 
It is not necessary that you multiply both sides by the least common multiple of the denominators. Any common multiple will do. The simplest way to do it is to multiply twice, once by each of the denominators (or, equivalently, multiply once by the product of the denominators). The drawback of this is that you then have more cancelling to do, and the cancelling can be nontrivial if you have multiplied by nonconstant factors such as (x2).

August 26th, 2015, 04:32 AM  #6 
Global Moderator Joined: Oct 2008 From: London, Ontario, Canada  The Forest City Posts: 7,697 Thanks: 977 Math Focus: Elementary mathematics and beyond 
It's not absolutely necessary, but it will make things easier, especially if the LCM is easily found (as is the case here). Ideally, I should have included a note saying so; point well taken.
Last edited by greg1313; August 26th, 2015 at 04:52 AM. 
August 26th, 2015, 07:41 AM  #7 
Math Team Joined: Jan 2015 From: Alabama Posts: 2,922 Thanks: 785 
After you have done all that, you will find that, as skipjack said, there are no solutions. Eventually, all "x" terms cancel, leaving "2= 3" which is NOT true, for any value of x.

August 26th, 2015, 11:48 AM  #8 
Global Moderator Joined: Oct 2008 From: London, Ontario, Canada  The Forest City Posts: 7,697 Thanks: 977 Math Focus: Elementary mathematics and beyond 
I was not assuming that interpretation of the question.
Last edited by greg1313; August 26th, 2015 at 11:54 AM. 

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