Question

How do you solve `3\leq \frac { 2- x } { x + 2}`?

Answer 1

The solution is `x in (-2,-1]`

Explanation:

We rewrite the inequality as

`3<=(2-x)/(x+2)`

`3-(2-x)/(x+2)<=0`

`(3(x+2)-(2-x))/(x+2)<=0`

`(3x+6-2+x)/(x+2)<=0`

`(4x+4)/(x+2)<=0`

`(4(x+1))/(x+2)<=0`

Let `f(x)=(4(x+1))/(x+2)`

We can build the sign chart

`color(white)(aaaa)` `x` `color(white)(aaaa)` `-oo` `color(white)(aaaa)` `-2` `color(white)(aaaaa)` `-1` `color(white)(aaaa)` `+oo`

`color(white)(aaaa)` `x+2` `color(white)(aaaaa)` `-` `color(white)(aa)` `||` `color(white)(aa)` `+` `color(white)(aaaaa)` `+`

`color(white)(aaaa)` `x+1` `color(white)(aaaaa)` `-` `color(white)(aaaaa)` `-` `color(white)(aa)` `0` `color(white)(aa)` `+`

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

Therefore,

`f(x)<=0` when `x in (-2,-1]`

graph{3-(2-x)/(x+2) [-28.87, 28.86, -14.43, 14.44]}