My Math Forum Adding algebra fractions

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 April 10th, 2014, 12:13 PM #1 Newbie   Joined: Apr 2014 From: UK Posts: 4 Thanks: 0 Adding algebra fractions Hello, i'm wracking my brains trying to figure this out. 1 _ 1 2x 3y So it's, (1 divided by 2x) subtract (1 divided by 3y) The problem is, how do I change the denominators so theyre the same. Any help would be amazing.
 April 10th, 2014, 01:05 PM #2 Senior Member   Joined: Jun 2013 From: London, England Posts: 1,316 Thanks: 116 Try 6xy as a common denominator.
 April 10th, 2014, 01:46 PM #3 Newbie   Joined: Apr 2014 From: UK Posts: 4 Thanks: 0 Really appreciate the reply. What would i change the numerator too? Would I change them like I changed the denominator ie multiply the first numerator by 3 and the second numerator by 2 and then just add them? Or would I find the LCM, then change the numerator accordingly, then multiply the new left numerator with the old right denominator and multiply the new right numerator with the old left denominator? I'm really confused by this question Again, thank you for the reply.
 April 10th, 2014, 02:06 PM #4 Newbie   Joined: Apr 2014 From: UK Posts: 4 Thanks: 0 i played around with a algebra calculator and finally got the answer 1/7*x+1/2*y Solution Steps 1 Simplify Multiply Terms 1/7x+1/2y 2 Eliminate Greatest Common Denominator 2y/14xy+7x/14xy 3 Simplify Fraction Addition 2y+7x/14xy 4 Order Terms by Power 7x+2y/14xy Appreciate the help though
 April 10th, 2014, 06:59 PM #5 Math Team   Joined: Dec 2013 From: Colombia Posts: 7,681 Thanks: 2659 Math Focus: Mainly analysis and algebra $$\frac{1}{2x}-\frac{1}{3y}$$ The simplest way to get a common denominator is to multiply the two denominators that you have. So in this case, the new denominator will be $6xy$. Then you have to make sure that, for each fraction, you multiply the top and bottom by the same number so that their values don't change. The denominator $2x$ is multiplied by $3y$ to get to $6xy$, so we have to multiply the numerator by $3y$ too. Similarly, the denominator $3y$ is multiplied by $2x$ to get to $6xy$, so we have to multiply the numerator by $2x$ too. This gives us $$\frac{1}{2x}-\frac{1}{3y} = \frac{3y}{6xy}-\frac{2x}{6xy}$$
 April 13th, 2014, 04:47 AM #6 Newbie   Joined: Apr 2014 From: UK Posts: 4 Thanks: 0 Thank you for the reply Pero and V8archie. I'm an independent student studying for my GCSE Maths exam, so I don't have a teacher to turn to. So your help means a lot!

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