How can you evaluate #((2x-5)(x-y)) / ((y-x)(3x-1))#?

1 Answer
Adrian D.
Jul 29, 2015

#-(2x-5)/(3x-1)#

Explanation:

First note that:
#((2x-5)(x-y))/((y-x)(3x-1))=-((2x-5)cancel((x-y)))/(cancel((x-y))(3x-1))#

So in fact this expression is only a function of #x# and the value of #y# is irrelevant. Plug the value of #x# into the remaining expression to evaluate it, for example #x=1#:

#-(2x-5)/(3x-1)=-(2-5)/(3-1)=-(-3)/(2)=3/2#

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