How do you solve #1/(x-1)+3/(x+1)=2#?

Question

4720 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-05-25T18:35:45

#x_1=0# and #x_2=2#

#1/(x-1)+3/(x+1)=2#

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

#(4x-2)/(x^2-1)=2#

#4x-2=2*(x^2-1)#

#x^2-1=2x-1#

#x^2-2x=0#

#x*(x-2)=0#

Hence #x_1=0# and #x_2=2#

Answer 2 · 2018-05-25T19:08:23

#x=0#, #x=2#

#1/(x-1)+3/(x+1)=2#

#rArr ((x+1)+3(x-1))/((x-1)(x+1))=2#

#rArr (x+1+3x-3)/((x-1)(x+1))=2#

#rArr (4x-2)/((x-1)(x+1))=2#

#rArr 4x-2=2(x+1)(x-1)#

#rArr 4x-2=2x+2(x-1)#

#rArr 4x-2=2x^2-2x+2x-2#

#rArr 4x-2=2x^2-2#

#rArr 2x^2-4x=0#

#rArr 2x(x-2)=0#

#rArr x-2=0 -> x=2#

#rArr 2x=0 -> x=0#