anonymous
  • anonymous
How to subtract a fraction with another fraction?
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
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anonymous
  • anonymous
[3/(x-1)] - [(2x+10)/(x^2+2x-3)] = 1/3
anonymous
  • anonymous
what is x?
anonymous
  • anonymous
this question is solved yours other subtittle.

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More answers

anonymous
  • anonymous
we say x=0
anonymous
  • anonymous
lcd = 3(x^2 +2x-3) = 3(x-1)(x+3) multiplying through by this gives: 9(x+3) - 3(2x+10) = (x-1)(x+3) 9x +27 - 6x - 30 = x^2 +2x - 3 x^2 +x = 0 x(x+1)=0 x = 0 or x = -1
anonymous
  • anonymous
yeah cnsu90, thanks for ur help :)
anonymous
  • anonymous
oops look like an error in last two steps! 9x +27 - 6x - 30 = x^2 +2x - 3 -x^2 +x = 0 x (1 - x) = 0 x = 0 or x = 1
anonymous
  • anonymous
hmm..
anonymous
  • anonymous
@jimmy..denominator cant be 0..so u cant say x=1..
anonymous
  • anonymous
yes - you are right
anonymous
  • anonymous
http://www.khanacademy.org/video/subtracting--fractions?playlist=Developmental+Math&exid=subtracting_fractions
anonymous
  • anonymous
jimmy, whats lcd? lowest common denominator?
anonymous
  • anonymous
yes
anonymous
  • anonymous
this is given me a real headache - cant understand how i got erroneous root
anonymous
  • anonymous
the anwser is x=0
anonymous
  • anonymous
yes - i know thats correct - i just trying to find out why my method also gave x=1
anonymous
  • anonymous
your solution is true jimmy.but then you must try your root in function.
anonymous
  • anonymous
there must be some reason for this
anonymous
  • anonymous
at x=1 function is undefined.
anonymous
  • anonymous
yea i know
anonymous
  • anonymous
- i just dont like mysteries...
anonymous
  • anonymous
ahh - i'd forgot a very important fact about multiplying the whole equation by the lcd!! - this will sometimes give extraneous roots - when x = 1 you are effectively multiplying by zero. you guys solved it differently - you simplified the lhs and left rhs alone - thats why you didn't get the extra root
anonymous
  • anonymous
its ok when you multiply by a constant - it only applies when you you multiply by a variable
anonymous
  • anonymous
hey jimmy, so which method is better? :)

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