Question

How do you solve `x/(x+2)>=2`?

Answer 1

The answer is `x in [-4, 2[`

Explanation:

You cannot do crossing over

`x/(x+2)>=2`

`x/(x+2)-2>=0`

`(x-2(x+2))/(x+2)>=0`

`(x-2x-4)/(x+2)>=0`

`(-x-4)/(x+2)>=0`

Let `f(x)=(-x-4)/(x+2)`

We can do a sign chart

`color(white)(aaaa)` `x` `color(white)(aaaaaa)` `-oo` `color(white)(aaaaa)` `-4` `color(white)(aaaaaaa)` `-2` `color(white)(aaaaa)` `+oo`

`color(white)(aaaa)` `-x-4` `color(white)(aaaaa)` `+` `color(white)(aaaa)` `0` `color(white)(aaa)` `-` `color(white)(aa)` `∥` `color(white)(aa)` `-`

`color(white)(aaaa)` `x+2` `color(white)(aaaaaaa)` `-` `color(white)(aaaa)` ###color(white)(aaa)#`-` `color(white)(aaa)` `∥` `color(white)(aa)` `+`

`color(white)(aaaa)` `f(x)` `color(white)(aaaaaaaa)` `-` `color(white)(aaaa)` ###color(white)(aaa)#`+` `color(white)(aaa)` `∥` `color(white)(aa)` `-`

Therefore,

`f(x)>=0`, when `x in [-4, 2[`