Question

How do you solve `m/(m-5)+3/(m-1)>0` using a sign chart?

Answer 1

The solution is `m in (-oo,-5)uu(1,3)uu(5,+oo)`

Explanation:

Let's do some simplification

`m/(m-5)+3/(m-1)>0`

`(m^2-m+3m-15)/((m-5)(m-1))>0`

`(m^2+2m-15)/((m-5)(m-1))>0`

`((m+5)(m-3))/((m-5)(m-1))>0`

Let `f(m)=((m+5)(m-3))/((m-5)(m-1))`

We can build the sign chart

`color(white)(aaaa)` `m` `color(white)(aaaa)` `-oo` `color(white)(aaaa)` `-5` `color(white)(aaaaaaa)` `1` `color(white)(aaaaa)` `3` `color(white)(aaaaaaa)` `5` `color(white)(aaaa)` `+oo`

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

`color(white)(aaaa)` `m-1` `color(white)(aaaa)` `-` `color(white)(aaaaa)` `-` `color(white)(aaaa)` `||` `color(white)(aa)` `+` `color(white)(aa)` `+` `color(white)(aaaa)` `||` `color(white)(aa)` `+`

`color(white)(aaaa)` `m-3` `color(white)(aaaa)` `-` `color(white)(aaaaa)` `-` `color(white)(aaaa)` `||` `color(white)(aa)` `-` `color(white)(aa)` `+` `color(white)(aaaa)` `||` `color(white)(aa)` `+`

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

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

So,

`f(m)>0` when `m in (-oo,-5)uu(1,3)uu(5,+oo)`