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

Answer 1

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

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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Answer 2

Eva Abramson

To evaluate the expression ((2x-5)(x-y)) / ((y-x)(3x-1)), we can simplify it by canceling out common factors.

First, let's factorize the expressions: (2x-5)(x-y) = 2x^2 - 2xy - 5x + 5y (y-x)(3x-1) = -3x^2 + x + 3xy - y

Now, we can rewrite the expression as: (2x^2 - 2xy - 5x + 5y) / (-3x^2 + x + 3xy - y)

Next, we can simplify further by canceling out common factors: (2x^2 - 2xy - 5x + 5y) / (-3x^2 + x + 3xy - y) = (x - y)(2x - 5) / (x - y)(-3x + 1)

Finally, we can cancel out the common factor of (x - y): (x - y)(2x - 5) / (x - y)(-3x + 1) = (2x - 5) / (-3x + 1)

Therefore, the simplified expression is (2x - 5) / (-3x + 1).

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.