Divide: 3a + 1 over a squared - 1 - 1 over a + 1?

Answer 1

assuming this is (3a+1)/(a^2-1) - 1/(a+1)

factorise the (a^2-1) to give (a+1)(a-1)

multiply the second fraction by (a-1)

--> (3a+1)/[(a+1)(a-1)] - 1(a-1)/[(a+1)(a-1)]

factorising this gives [(3a+1)-(a-1)]/[(a+1)(a-1)]

which equals (2a+2)/[(a+1)(a-1)]

= [2(a+1)]/[(a+1)(a-1)]

the (a+1)'s cancel out...

therefore = 2/(a-1)

Answer 2

((3a + 1)/(a^2 - 1)) - (1/(a + 1))
((3a + 1)/((a - 1)(a + 1))) - (1/(a + 1))

multiply everything by (a - 1)

((3a + 1) - (a - 1))/((a - 1)(a + 1))
(3a + 1 - a + 1)/((a - 1)(a + 1))
(2a + 2)/((a - 1)(a + 1))
(2(a + 1))/((a - 1)(a + 1))

ANS : 2/(a - 1)