Question

How do you solve `(2x)/(x-5)>=3`?

Answer 1

The solution is `x in (5,15]`

Explanation:

You cannot do crossing over in inequalities

`(2x)/(x-5)>=3`

`<=>`, `(2x)/(x-5)-3>=0`

`<=>`, `(2x-3(x-5))/(x-5)>=0`

`<=>`, `(2x-3x+15)/(x-5)>=0`

`<=>`, `(-x+15)/(x-5)>=0`

Let `f(x)=(-x+15)/(x-5)`

Build a sign chart

`color(white)(aaaa)` `x` `color(white)(aaaa)` `-oo` `color(white)(aaaaaaa)` `5` `color(white)(aaaaaaa)` `15` `color(white)(aaaa)` `+oo`

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

`color(white)(aaaa)` `-x+15` `color(white)(aaa)` `+` `color(white)(aaaaa)` ###color(white)(aa)#`+` `color(white)(aaa)` `0` `color(white)(aa)` `-`

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

Therefore,

`f(x)>=0` when `x in (5,15]`

graph{(2x)/(x-5)-3 [-25.56, 32.17, -14.65, 14.22]}