Question

Question #e97c3

Answer 1

The solution is `x in ]-oo,1[ uu [4,+oo[`

Explanation:

Let's simplify the equation

`(2x+1)/(x-1)<=3`

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

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

`(4-x)/(x-1) <=0`

Let `f(x)=(4-x)/(x-1)`

We can build the sign chart

`color(white)(aaaa)` `x` `color(white)(aaaa)` `-oo` `color(white)(aaaa)` `1` `color(white)(aaaaaaaa)` `4` `color(white)(aaaa)` `+oo`

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

`color(white)(aaaa)` `4-x` `color(white)(aaaa)` `+` `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 ]-oo,1[ uu [4,+oo[`