Question

Question #bfc27

Answer 1

The solution is `x in (1, 10/3)`

Explanation:

We cannot do crossing over

So, let's rearrange the inequality

`(2x+5)/(x-1)>5`

`(2x+5)/(x-1)-5>0`

`((2x+5)-5(x-1))/(x-1)>0`

`(2x+5-5x+5)/(x-1)>0`

`(10-3x)/(x-1)>0`

Let `f(x)=(10-3x)/(x-1)`

We can build a sign chart

`color(white)(aaaa)` `x` `color(white)(aaaa)` `-oo` `color(white)(aaaa)` `1` `color(white)(aaaaaaa)` `10/3` `color(white)(aaaa)` `+oo`

`color(white)(aaaa)` `x-1` `color(white)(aaaa)` `-` `color(white)(aa)` `||` `color(white)(aaaa)` `+` `color(white)(aaaa)` `+`

`color(white)(aaaa)` `10-3x` `color(white)(aa)` `+` `color(white)(aa)` `||` `color(white)(aaaa)` `+` `color(white)(aaaa)` `-`

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

Therefore,

`f(x)>0` when `x in (1, 10/3)`