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

2 Answers
Feb 20, 2017

Answer:

#x<-3#

Explanation:

#1/(x+1)>2/(x-1) #

#1/(x+1)-2/(x-1)>0#

#((x-1)-2(x+1))/(x^2-1)>0# [ #x!=+-1#]

#-x-3>0 -> x+3<0##

#:. x<-3#

Feb 20, 2017

Answer:

The solution is #x in ]-oo, -3[uu]-1,1[#

Explanation:

We cannot do crossing over

Let's rearrange the equation

#1/(x+1)>2/(x-1)#

#2/(x-1)-1/(x+1)<0#

#(2x+2-x+1)/((x+1)(x-1))<0#

#(x+3)/((x+1)(x-1))<0#

Let #f(x)=(x+3)/((x+1)(x-1))#

We can build the sign chart

#color(white)(aaaa)##x##color(white)(aaaa)##-oo##color(white)(aaaa)##-3##color(white)(aaaaaaa)##-1##color(white)(aaaaaaa)##1##color(white)(aaaaaaa)##+oo#

#color(white)(aaaa)##x+3##color(white)(aaaaa)##-##color(white)(aaaaa)##+##color(white)(aaaa)##||##color(white)(aa)##+##color(white)(aaa)##||##color(white)(aaaa)##+#

#color(white)(aaaa)##x+1##color(white)(aaaaa)##-##color(white)(aaaaa)##-##color(white)(aaaa)##||##color(white)(aa)##+##color(white)(aaa)##||##color(white)(aaaa)##+#

#color(white)(aaaa)##x-1##color(white)(aaaaa)##-##color(white)(aaaaa)##-##color(white)(aaaa)##||##color(white)(aa)##-##color(white)(aaa)##||##color(white)(aaaa)##+#

#color(white)(aaaa)##f(x)##color(white)(aaaaaa)##-##color(white)(aaaaa)##+##color(white)(aaaa)##||##color(white)(aa)##-##color(white)(aaa)##||##color(white)(aaaa)##+#

Therefore,

#f(x)<0# when #x in ]-oo, -3[uu]-1,1[#