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 August 24th, 2015, 06:29 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(2x-1) = 1/2(3x+1)
 August 24th, 2015, 07:19 PM #2 Global Moderator     Joined: Oct 2008 From: London, Ontario, Canada - The Forest City Posts: 7,599 Thanks: 941 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 Thanks from Denis and Benit13
 August 25th, 2015, 05:02 AM #3 Math Team   Joined: Jan 2015 From: Alabama Posts: 2,734 Thanks: 707 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, 05:06 AM #4 Global Moderator   Joined: Dec 2006 Posts: 18,059 Thanks: 1396 $\displaystyle \frac{1}{3(2x- 1)} = \frac{1}{2(3x- 1)}$ has no solutions.
 August 25th, 2015, 05:21 AM #5 Math Team   Joined: Dec 2013 From: Colombia Posts: 6,973 Thanks: 2296 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 non-trivial if you have multiplied by non-constant factors such as (x-2).
 August 26th, 2015, 03:32 AM #6 Global Moderator     Joined: Oct 2008 From: London, Ontario, Canada - The Forest City Posts: 7,599 Thanks: 941 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 03:52 AM.
 August 26th, 2015, 06:41 AM #7 Math Team   Joined: Jan 2015 From: Alabama Posts: 2,734 Thanks: 707 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, 10:48 AM #8 Global Moderator     Joined: Oct 2008 From: London, Ontario, Canada - The Forest City Posts: 7,599 Thanks: 941 Math Focus: Elementary mathematics and beyond I was not assuming that interpretation of the question. Last edited by greg1313; August 26th, 2015 at 10:54 AM.

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