How do you combine $(x-1)*[3/(x^2-1)+x/(2x-2)]$ ?

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Answer 1 · 2017-08-22T22:29:25

$=(x^2+x+6)/(2(x+1))$

$(x-1)*[3/(x^2-1)+x/(2x-2)]$

Factorise the denominators first

$= (x-1)*[3/((x+1)(x-1))+x/(2(x-1))]$

Add the fractions using a common denominator.

$=(x-1)*[((2xx3 + x(x+1))/(2(x+1)(x-1))]$

$=(x-1)*[(6 + x^2+x)/(2(x+1)(x-1))]$

Distribute $(x-1)$ into the bracket:

$=[(cancel((x-1))(6 + x^2+x))/(2(x+1)cancel((x-1))]$

$=(x^2+x+6)/(2(x+1))$

How do you combine (x-1)*[3/(x^2-1)+x/(2x-2)](x-1)*[3/(x^2-1)+x/(2x-2)]?