Question

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

Answer 1

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)`

Answer 2

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]}