Question

How do you solve `4/(x-3)<2` using a sign chart?

Answer 1

The answer is `x in ]-oo, 3[uu]5, +oo[`

Explanation:

Let's rearrange the equation

`4/(x-3)-2<0`

`(4-2(x-3))/(x-3)<0`

`(4-2x+6)/(x-3)<0`

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

Let `f(x)=(10-2x)/(x-3)`

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

Now, we build the sign chart

`color(white)(aaaa)` `x` `color(white)(aaaaa)` `-oo` `color(white)(aaaaa)` `3` `color(white)(aaaaaaa)` `5` `color(white)(aaaaaa)` `+oo`

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

`color(white)(aaaa)` `10-2x` `color(white)(aaaa)` `+` `color(white)(aa)` `||` `color(white)(aaaa)` `+` `color(white)(aaaa)` `-`

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

Therefore,

`f(x)<0` when `x in ]-oo, 3[uu]5, +oo[`