Question

How do you solve and write the following in interval notation: `3/(x-1) - 4/x>=1`?

Answer 1

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

Explanation:

Let's rearrange the inequality

`3/(x-1)-4/x>=1`

`3/(x-1)-4/x-1>=0`

`(3x-4(x-1)-x(x-1))/(x(x-1))>=0`

`(3x-4x+4-x^2+x)/(x(x-1))>=0`

`(4-x^2)/(x(x-1))>=0`

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

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

We can build the sign chart

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

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

`color(white)(aaaa)` `x` `color(white)(aaaaaaaa)` `-` `color(white)(aa)` ###color(white)(aaaa)#`-` `color(white)(a)` `||` `color(white)(a)` `+` `color(white)(aaa)` `+` `color(white)(aaa)` `+`

`color(white)(aaaa)` `x-1` `color(white)(aaaaa)` `-` `color(white)(aa)` ###color(white)(aaaa)#`-` `color(white)(a)` ###color(white)(aa)#`-` `color(white)(a)` `||` `color(white)(a)` `+` `color(white)(aaa)` `+`

`color(white)(aaaa)` `2-x` `color(white)(aaaaa)` `+` `color(white)(aa)` ###color(white)(aaaa)#`+` `color(white)(a)` ###color(white)(aa)#`+` `color(white)(a)` ###color(white)(aa)#`+` `color(white)(a)` `0` `color(white)(a)` `-`

`color(white)(aaaa)` `f(x)` `color(white)(aaaaaa)` `-` `color(white)(aa)` `0` `color(white)(aaa)` `+` `color(white)(a)` `||` `color(white)(a)` `-` `color(white)(a)` `||` `color(white)(a)` `+` `color(white)(a)` `0` `color(white)(a)` `-`

Therefore,

`f(x)>=0` when `x in [-2,0) uu (1, 2]`

graph{3/(x-1)-4/x-1 [-35.57, 37.5, -25.57, 10.98]}