Question

How do you solve `(3x)/(x + 2)<5`?

Answer 1

The solution is `x in (-oo,-5) uu (-2, +oo)`

Explanation:

We cannot do crossing over, we rearrange the inequality

`(3x)/(x+2)<5`

`(3x)/(x+2)-5<0`

`(3x-5(x+2))/(x+2)<0`

`(3x-5x-10)/(x+2)<0`

`(-2x-10)/(x+2)<0`

`(-2(x+5))/(x+2)<0`

Let `f(x)=(-2(x+5))/(x+2)`

Let's build a sign chart

`color(white)(aaaa)` `x` `color(white)(aaaaaaa)` `-oo` `color(white)(aaaa)` `-5` `color(white)(aaaaaaa)` `-2` `color(white)(aaaaaaa)` `+oo`

`color(white)(aaaa)` `-(x+5)` `color(white)(aaaaaa)` `+` `color(white)(aaaa)` `-` `color(white)(aaaa)` `||` `color(white)(aaaa)` `-`

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

`color(white)(aaaa)` `f(x)` `color(white)(aaaaaaaaaaa)` `-` `color(white)(aaaa)` `+` `color(white)(aaaa)` `||` `color(white)(aaaa)` `-`

Therefore,

`f(x)<0`, when `x in (-oo,-5) uu (-2, +oo)`
graph{(3x)/(x+2)-5 [-36.53, 36.52, -18.28, 18.27]}