How do you combine #x^2/(x-1) - 1/(x-1)#?

2 Answers
May 17, 2015

Since the numerators are the same:

#x^2/(x-1) - 1/(x-1) = (x^2-1)/(x-1)#

Now #x^2 - 1 = (x+1)(x-1)#

So #(x^2-1)/(x-1) = ((x+1)(x-1))/(x-1) = x+1#

May 17, 2015

You do it by using the lowest common denominator among both fractions.

However, when both fractions have the same denominator and you're working with sums or subtractions, you can just go directly to the operation, summing/subtracting the terms in the numerators, keeping the denominator as it is, as follows:

#(x^2)/(x-1)-1/(x-1) = (x^2-1)/(x-1)#

Note that the numerator #x^2-1# has two roots, which we can find by isolating #x#:

#x^2=1#
#x=sqrt(1)#
#x=1# and #x=-1#, which is the same as if we wrote:
#x-1=0# and #x+1=0#.

By factoring #x^2-1# using its two roots, we can have your answer like this:

#(cancel((x-1))(x+1))/cancel(x-1) = x+1#

So, your final answer is #x+1#