Question

How do you solve `(3x)/(x^2+3x)>=1` ?

`(3x)/(x^2+3x)>=1 => (3x-x(x+3))/(x(x+3))>=0 => (-x^2)/(x(x+3))>=0 *-1 => (x^2)/(x(x+3))<=0`
`x^2=0 => x=0`
`x=0`
`x+3=0 => x=-3`

`x=0 (2k)`
Answer:
`(-oo;-3)`

Answer 1

The solution is `x in (-3,0]`

Explanation:

The first step is ok

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

Divide by `x`

`=>`, `x/(x+3)<=0`

Let `f(x)=x/(x+3)`

Make a sign chart

`color(white)(aaaa)` `x` `color(white)(aaaa)` `-oo` `color(white)(aaaaa)` `-3` `color(white)(aaaaaaaa)` `0` `color(white)(aaaaaaaaa)` `+oo`

`color(white)(aaaa)` `x` `color(white)(aaaaaaaaa)` `-` `color(white)(aaaaaa)` `-` `color(white)(aaaa)` `0` `color(white)(aaaa)` `+`

`color(white)(aaaa)` `x+3` `color(white)(aaaaa)` `-` `color(white)(aaa)` `||` `color(white)(aaa)` `+` `color(white)(aaaaaaaa)` `+`

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

Therefore,

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

graph{x/(x+3) [-20.28, 20.27, -10.14, 10.14]}