Question 1069303
{{{1+2/(x+1)<=2x}}} <--> {{{2/(x+1)<=2x-1}}} 
After that we have to think.
We know that {{{x+1=0}}} <--> {{{x=-1}}} is forbidden.
so, either {{{x+1>0}}} <--> {{{x>-1}}} ,
or {{{x+1>0}}} <--> {{{x>-1}}} .
We also know that if we multiply both sides of an inequality times a positive number
(or divide by a positive number),
the relationship (greater than, or lesser than) remains the same,
but if we multiply times or divide by a negative number,
the resulting inequality has the opposite sign.
So, for {{{x+1>0}}} <--> {{{x>-1}}} ,
{{{2/(x+1)<=2x-1}}} is equivalent to
{{{2<=(2x-1)(x+1)}}}<-->{{{0<=(2x-1)(x+1)-2}}}<-->{{{0<=(2x^2+x-1)-2}}}<-->{{{2x^2+x-3>=0}}} ,
and for {{{x+1<0}}} <--> {{{x<-1}}} ,
{{{2/(x+1)<=2x-1}}} is equivalent to
{{{2>=(2x-1)(x+1)}}}<-->{{{0>=(2x-1)(x+1)-2}}}<-->{{{0>=(2x^2+x-1)-2}}}<-->{{{2x^2+x-3<=0}}} .
So, the solutions to {{{1+2/(x+1)<=2x}}} will be
the solutions to {{{system(x>-1,2x^2+x-3>=0)}}}
and the solutions to {{{system(x<-1,2x^2+x-3<=0)}}} .
We can factor {{{2x^2+x-3}}} ,
which is better than having to find the zeros any other way,
makes it easier to see when when it is positive, negative, or zero.
{{{2x^2+x-3=(x-1)(2x+3)=0}}} for {{{x=-3/2=-1.5}}} and for {{{x=1}}} .
{{{2x^2+x-3=(x-1)(2x+3)<=0}}} for {{{-3/2=-1.5<=x<=1}}} (in between the zeros).
{{{2x^2+x-3=(x-1)(2x+3)>=0}}} for {{{x<-3/2=-1.5}}} or for {{{x>=1}}} .
The solutions to {{{system(x<-1,2x^2+x-3<=0)}}}
are the solutions to {{{system(x<-1,-3/2=-1.5<=x<=1)}}} ,
and that is {{{-3/2=-1.5<x<-1}}} or {{{highlight("[ -1.5 , -1 )")}}} .
The solutions to {{{system(x>-1,2x^2+x-3>=0)}}}
are the solutions to {{{system(x>-1,x>=1)}}} ,
and that is {{{x>=1}}} or {{{highlight(matrix(1,3,"[ 1 ,",infinity,")"))}}} .