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anonymous
 4 years ago
How to subtract a fraction with another fraction?
anonymous
 4 years ago
How to subtract a fraction with another fraction?

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anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0[3/(x1)]  [(2x+10)/(x^2+2x3)] = 1/3

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0this question is solved yours other subtittle.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0lcd = 3(x^2 +2x3) = 3(x1)(x+3) multiplying through by this gives: 9(x+3)  3(2x+10) = (x1)(x+3) 9x +27  6x  30 = x^2 +2x  3 x^2 +x = 0 x(x+1)=0 x = 0 or x = 1

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0yeah cnsu90, thanks for ur help :)

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0oops 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
 4 years ago
Best ResponseYou've already chosen the best response.0@jimmy..denominator cant be 0..so u cant say x=1..

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0jimmy, whats lcd? lowest common denominator?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0this is given me a real headache  cant understand how i got erroneous root

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0yes  i know thats correct  i just trying to find out why my method also gave x=1

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0your solution is true jimmy.but then you must try your root in function.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0there must be some reason for this

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0at x=1 function is undefined.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0 i just dont like mysteries...

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0ahh  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
 4 years ago
Best ResponseYou've already chosen the best response.0its ok when you multiply by a constant  it only applies when you you multiply by a variable

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0hey jimmy, so which method is better? :)
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