Simplifying Fractions?

  • Thread starter wocket92
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  • #1
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Hey, I need some help with simplifying these fractions.

(3x-6)/(x^2-3x+2) - (x^2-1)/(x^2-4x+4)

Im not quite sure what to do with the denominators. Am I supposed to change them to special forms? What is the process of making these a single fraction?
Thanks in advance.
 
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Answers and Replies

  • #2
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I would love to help you but It’s nearly impossible to tell what you mean!
3x-6/x^2-3x+2 - x^2-1/x^2-4x+4 could mean a ton of things, for example:

(3x-6/x^2)-3x+2 –( x^2-1)/(x^2-4x+4)
or
3x- (6/x^2) -3x+2 - x^2- (1/x^2) -4x+4

If I had to guess I would suspect you mean:
(3x-6)/(x^2-3x+2) – (x^2-1)/(x^2-4x+4)

If so factor the numerator and the denominator of each of these fractions.
 
  • #3
HallsofIvy
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Hey, I need some help with simplifying these fractions.

(3x-6)/(x^2-3x+2) - (x^2-1)/(x^2-4x+4)

Im not quite sure what to do with the denominators. Am I supposed to change them to special forms? What is the process of making these a single fraction?
Thanks in advance.
In order to add or subtract fractions, in algebra as in arithmetic, you must get "common denominators". And, again, as in arithmetic, you can get the least common denominator by factoring the denominators.

Here, [itex]x^2- 3x+ 2= (x- 2)(x- 1)[/itex] and [itex](x^2- 4x+ 4)= (x- 2)^2[/itex]. The "least common denominator would be [itex](x- 2)^2(x- 1)[/itex]. You can get that by multiplying both numerator and denominator of the first fraction by x- 2 and multiplying both numerator and denominator of the second fraction by x- 1.
 

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