How do you solve # [a / (a+1)] - [1 / (a-1)]#?

1 Answer
Nov 6, 2015

Answer:

#(a^2-2a-1)/(a^2-1)#

Look at the method. It is another technique for you to consider.

Explanation:

#color(blue)("Things to think about for the method")#
These are fraction # -> (count)/(size)#

#color(brown)("To be able to directly add or subtract counts")# what you are counting #color(green)("must be of the same type or size.")#

#(count)/(size) -> ("numerator")/("denominator")#

Neither of #(a+1) # nor #(a-1)# will easily divide into each other. So you have to devise something they will. The easiest way to do this is multiply them together.

Write #(a+1) times (a-1)# as #(a+1)(a-1)#

Consider #a/(a+1)#. How do we change #(a+1)# into #(a+1)(a-1)#?

#color(green)("If we multiply it by 1 but in the form of "(a-1)/(a-1)")# we change the way it looks but not its actual value. Think of #2 times 3/3 = 6/3#.

~~~~~~~~~~~~~~~~~~~~~~ end of method introduction ~~~~~~~

#[a/(a+1) times (a-1)/(a-1)] -[1/(a-1) times (a+1)/(a+1)]#

#[(a(a-1))/((a+1)(a-1))] - [(a+1)/((a+1)(a-1))]#

#((a^2-a) -(a+1))/((a+1)(a-1))#

#(a^2-2a-1)/((a+1)(a-1))#

but #a^2-1^2 = a^2-1 =(a+1)(a-1)#

So by substitution we have:

#(a^2-2a-1)/(a^2-1)#