evaluate 2-(x+1/x-1):3x^2+3x/6x^2

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You need to evaluate:  `2- (x+1)/(x-1):(3x^2+3x)/(6x^2)`

You need to use the property of division of two normal fractions hence you need to multiply `(x+1)/(x-1)`  by reciprocal of `(3x^2+3x)/(6x^2)`  such that:

`(x+1)/(x-1):(3x^2+3x)/(6x^2) = (x+1)/(x-1)*(6x^2)/(3x^2+3x)`

Factoring out 3x to denominator of reciprocal yields:


Reducing by x+1 yields: `(6x^2)/(3x(x-1))`

Reducing by 3x yields: `(2x)/(x-1).`

There is no more to reduce, hence you may subtract `(2x)/(x-1) ` form 2.

`2 - (2x)/(x-1)`

Bringing the terms to a common denominator yields:

`(2(x - 1) - 2x)/(x - 1) = (2x - 2 - 2x)/(x - 1)`

`` Reducing like terms to numerator yields: `-2/(x-1).`

Hence, evaluating the difference yields a negative normal fraction `-2/(x-1).`