Question

How do you solve `3/(x-2)<=3/(x+3)` using a sign chart?

Answer 1

The solution is `x in ]-3,2[`

Explanation:

We cannot do crossing over, so rewrite the inequality

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

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

Placing on the same denominator

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

`(3x+9-3x+6)/((x-2)(x+3))<=0`

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

Let `f(x)=15/((x-2)(x+3))`

The domain of `f(x)` is `D_f(x)=RR-{-3,2}`

Now, we can build the sign chart

`color(white)(aaaa)` `x` `color(white)(aaaa)` `-oo` `color(white)(aaaaa)` `-3` `color(white)(aaaaaa)` `2` `color(white)(aaaaaaa)` `+oo`

`color(white)(aaaa)` `x+3` `color(white)(aaaa)` `-` `color(white)(aaaa)` `||` `color(white)(aa)` `+` `color(white)(aa)` `||` `color(white)(aaaa)` `+`

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

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

Therefore,

`f(x)<=0` when `x in ]-3,2[`