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April 2nd, 2018, 03:51 PM  #1 
Newbie Joined: Mar 2018 From: Canada Posts: 14 Thanks: 0  Least common denominator
Hi everyone. The following is giving me some trouble. (x^2 + 2x)/(x+2) + (x)/(x1) I’m supposed to find the LCD. Here is what is giving me trouble. Do I have to first factor the numerators and then find LCD? ( that would give me (x1) as LCD) If I don’t factor the numerators first, I believe I would get (x1).(x+2) as my answer? So my question is? Are both ways of doing it correct? Or should I actually always factor numerators first? Can this expression have 2 LCD? Or it should always have just one ? Thank you very much! Last edited by Brunob; April 2nd, 2018 at 03:54 PM. 
April 2nd, 2018, 06:02 PM  #2 
Senior Member Joined: May 2016 From: USA Posts: 1,310 Thanks: 552 
$\dfrac{x^2 + 2x}{x + 2} + \dfrac{x}{x  1} = \dfrac{(x^2 + 2x)(x  1)}{(x + 2)(x  1)} + \dfrac{(x + 2)x}{(x + 2)(x  1)} =$ $\dfrac{x^3  x^2 + 2x^2  2x + x^2 + 2x}{(x + 2)(x  1)} = \dfrac{x^2(x + 2)}{(x + 2)(x  1)} = \dfrac{x^2}{x  1}.$ ALTERNATIVELY. $\dfrac{x^2 + 2x}{x + 2} + \dfrac{x}{x  1} = \dfrac{x(x + 2)}{x + 2} + \dfrac{x}{x  1} =$ $x + \dfrac{x}{x  1} = \dfrac{x(x  1)}{x  1} + \dfrac{x}{x  1} = \dfrac{x^2  x + x}{x  1} = \dfrac{x^2}{x  1}.$ It makes absolutely no difference to the result which way you do it. They are equivalent. It is usually easier to simplify the fractions first if possible. 
April 2nd, 2018, 06:48 PM  #3  
Math Team Joined: Oct 2011 From: Ottawa Ontario, Canada Posts: 14,597 Thanks: 1039  Quote:
have known that by assigning a value to x, like x=5; then: (x^2 + 2x)/(x+2) + (x)/(x1) = 6.25 (x1).(x+2) = 28 Doing this during simplification tells you if you're on the right path or not...get it, tabarnak?!  
April 2nd, 2018, 08:22 PM  #4 
Newbie Joined: Nov 2013 Posts: 28 Thanks: 8  Denis, You are mistaken. The OP just asked for the LCD and claimed the answer (x1).(x+2) is correct. As a result your substitution was not necessary. So what do you have to say about them apples?

April 2nd, 2018, 08:26 PM  #5  
Newbie Joined: Nov 2013 Posts: 28 Thanks: 8  Quote:
Consider this problem: 2(x3)/(x3) + 3(x+7)/(x+7). How would you do this one? I would just do 2+3=5 and be done. Last edited by Jomo; April 2nd, 2018 at 08:29 PM.  
April 3rd, 2018, 05:10 AM  #6  
Math Team Joined: Oct 2011 From: Ottawa Ontario, Canada Posts: 14,597 Thanks: 1039  Quote:
I answered a different question I'll go stand in the corner for (x1)(x+2) minutes... Thanks for the reprimand, Oh Jomo  
April 3rd, 2018, 05:11 PM  #7 
Newbie Joined: Nov 2013 Posts: 28 Thanks: 8  
April 3rd, 2018, 07:00 PM  #8 
Math Team Joined: May 2013 From: The Astral plane Posts: 2,340 Thanks: 983 Math Focus: Wibbly wobbly timeywimey stuff.  
April 3rd, 2018, 08:36 PM  #9  
Senior Member Joined: May 2016 From: USA Posts: 1,310 Thanks: 552  Quote:
It is, I think, a very stupid question. I suspect that what was meant was the simplest common denominator. In any case, I carefully avoided answering it. Last edited by JeffM1; April 3rd, 2018 at 08:39 PM.  
April 4th, 2018, 07:08 PM  #10 
Newbie Joined: Mar 2018 From: Canada Posts: 14 Thanks: 0 
Thanks everyone!


Tags 
algebra precalculus, common, denominator, lcd 
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