Question

How do you solve `\frac { x + 1} { x - 6} - \frac { x - 1} { x + 6} \leq 0`?

Answer 1

The solution is `x in (-oo,-6) uu [0, 6)`

Explanation:

We rearrange and simplify the inequality

`(x+1)/(x-6)-(x-1)/(x+6)<=0`

`((x+1)(x+6)-(x-1)(x-6))/((x+6)(x-6))<=0`

`((x^2+7x+6)-(x^2-7x+6))/((x+6)(x-6))<=0`

`((x^2+7x+6-x^2+7x-6))/((x+6)(x-6))<=0`

`((14x))/((x+6)(x-6))<=0`

Let `f(x)=((14x))/((x+6)(x-6))`

We can build the sign chart

`color(white)(aaaa)` `x` `color(white)(aaaa)` `-oo` `color(white)(aaaaaaa)` `-6` `color(white)(aaaaa)` `0` `color(white)(aaaaaa)` `6` `color(white)(aaaaa)` `+oo`

`color(white)(aaaa)` `x+6` `color(white)(aaaaaa)` `-` `color(white)(aaaa)` `||` `color(white)(aa)` `+` `color(white)(aaaa)` `+` `color(white)(aaaa)` `+`

`color(white)(aaaa)` `x` `color(white)(aaaaaaaaa)` `-` `color(white)(aaaa)` `color(white)(aaa)` `-` `color(white)(aa)` `0` `color(white)(aa)` `+` # color(white)(aaaa)# `+`

`color(white)(aaaa)` `x-6` `color(white)(aaaaaa)` `-` `color(white)(aaaaaaa)` `-` `color(white)(aaaa)` `-` `color(white)(a)` `||` `color(white)(aaa)` `+`

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

Therefore,

`f(x)<=0` when `x in (-oo,-6) uu [0, 6)`

graph{(x+1)/(x-6)-(x-1)/(x+6) [-41.1, 41.1, -20.56, 20.56]}