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
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}
$
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!
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