simplify

$\displaystyle \frac {x-3}{x^2-3x+2} - \frac {x-2}{x^2-4x+3}$

my answer

$\displaystyle

\frac {(x-3)(x-2)}{(x-2)^2}

$

correct? thanks

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- Nov 1st 2008, 12:20 AMjvignaciosimplify check!
simplify

$\displaystyle \frac {x-3}{x^2-3x+2} - \frac {x-2}{x^2-4x+3}$

my answer

$\displaystyle

\frac {(x-3)(x-2)}{(x-2)^2}

$

correct? thanks - Nov 1st 2008, 12:56 AMCaptainBlack
- Nov 1st 2008, 01:01 AMjvignacio
sorry mate. ok here it is:

$\displaystyle \frac {x-3}{x^2-3x+2} - \frac {x-2}{x^2-4x+3}$

$\displaystyle \frac {x-3}{(x-2)(x-1)} - \frac {x-2}{(x-3)(x-1)}$

then cross multiply then mulitply denominators and start cancelling out.

$\displaystyle

\frac {(x-3)(x-2)}{(x-1)^2}

$ - Nov 1st 2008, 01:40 AMCaptainBlack
- Nov 1st 2008, 01:48 AMjvignacio
- Nov 1st 2008, 01:56 AMmr fantastic
- Nov 1st 2008, 02:06 AMjvignacio
- Nov 1st 2008, 03:08 AMmr fantastic
NO!

Do you really think that $\displaystyle \frac{3 \times 5 - 2}{3}$ is equal to 3!! I was trying to make a point. The point is that you can't do this. What you cancel has to be common to**all of the numerator**and**all of the denominator**before it can be cancelled! - Nov 1st 2008, 04:07 AMCaptainBlack

You can only cancel something which is a factor of both the top and bottom of such an expressions and $\displaystyle x-3$ is not a factor of the top.

What you do is expand the brackets at the top, collect like powers and factorise. Then if any of the factors on top are the same as a factor on the bottom you can canncel such facors.

CB - Nov 1st 2008, 08:40 PMjvignacio
- Nov 1st 2008, 08:45 PMmr fantastic