Question

How do you solve `5x - 1\leq \frac { 2x - 1} { x + 1}`?

Answer 1

The solution is `x in (-oo,-1) uu [-2/5,0]`

Explanation:

We cannot do crossing over, we simplify the inequality

`5x-1<=(2x-1)/(x+1)`

`(5x-1)-(2x-1)/(x+1)<=0`

`((5x-1)(x+1)-(2x-1))/(x+1)<=0`

`(5x^2+4x-1-2x+1)/(x+1)<=0`

`(5x^2+2x)/(x+1)<=0`

`(x(5x+2))/(x+1)<=0`

Let `f(x)=(x(5x+2))/(x+1)`

Now, we build the sign chart

`color(white)(aaaa)` `x` `color(white)(aaaa)` `-oo` `color(white)(aaaaa)` `-1` `color(white)(aaaaaa)` `-2/5` `color(white)(aaaaaa)` `0` `color(white)(aaaa)` `+oo`

`color(white)(aaaa)` `x+1` `color(white)(aaaa)` `-` `color(white)(aaaa)` `||` `color(white)(aa)` `+` `color(white)(aaaa)` `0` `color(white)(aa)` `+` `color(white)(aa)` `0` `color(white)(aa)` `+`

`color(white)(aaaa)` `5x+2` `color(white)(aaa)` `-` `color(white)(aaaa)` `||` `color(white)(aa)` `-` `color(white)(aaaa)` `0` `color(white)(aa)` `+` `color(white)(aa)` `0` `color(white)(aa)` `+`

`color(white)(aaaa)` `x` `color(white)(aaaaaaaa)` `-` `color(white)(aaaa)` `||` `color(white)(aa)` `-` `color(white)(aaaa)` `0` `color(white)(aa)` `-` `color(white)(aa)` `0` `color(white)(aa)` `+`

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

Therefore,

`f(x)<=0` when `x in (-oo,-1) uu [-2/5,0]`

graph{(5x^2+2x)/(x+1) [-50.9, 53.13, -30.16, 21.9]}

graph{(5x^2+2x)/(x+1) [-50.9, 53.13, -30.16, 21.9]}