How do you solve and write the following in interval notation: #(x+1) / (x-1) <=2#?

2 Answers

Answer:

the interval is #[3,oo)#

Explanation:

the given inequlity is #(x+1)/(x-1)<=2# #rArr(x+1)<=2(x-1)rArr(x+1)<=2x-2# #rArr(x-2x+1+2)<=0rArr(-x+3)<=0# #rArr(x-3)>=0rArrx>=3#
#:.x in [3,oo)#

Nov 1, 2017

Answer:

The solution is #x in (-oo,1) uu [3,+oo)#

Explanation:

We cannot do crossing over

Let's rewrite the expression

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

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

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

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

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

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

Let's build a sign chart

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

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

#color(white)(aaaa)##3-x##color(white)(aaaaa)##+##color(white)(aaaa)##||##color(white)(aa)##+##color(white)(aa)##0##color(white)(aaaa)##-#

#color(white)(aaaa)##f(x)##color(white)(aaaaaa)##-##color(white)(aaaa)##||##color(white)(aa)##+##color(white)(aa)##0##color(white)(aaaa)##-#

Therefore,

#f(x)>=0# when #x in (-oo,1) uu [3,+oo)#

graph{(x+1)/(x-1)-2 [-12.66, 12.65, -6.33, 6.33]}