Question

How do you solve and write the following in interval notation: `12/ (x-5) > 10 /(x+1)`?

Answer 1

The solution is `x in (-31,-1) uu (5, +oo)`

Explanation:

We cannot do crossing over.

Rearrange the inequality

`12/(x-5)>10/(x+1)`

`12/(x-5)-10/(x+1)>0`

`(12(x+1)-10(x-5))/((x-5)(x+1))>0`

`(12x+12-10x+50)/((x-5)(x+1))>0`

`(2x+62)/((x-5)(x+1))>0`

`(2(x+31))/((x-5)(x+1))>0`

Let `f(x)=(2(x+31))/((x-5)(x+1))`

We can build a sign chart

`color(white)(aaaa)` `x` `color(white)(aaaa)` `-oo` `color(white)(aaaa)` `-31` `color(white)(aaaaaa)` `-1` `color(white)(aaaaaaaa)` `5` `color(white)(aaaa)` `+oo`

`color(white)(aaaa)` `x+31` `color(white)(aaaaa)` `-` `color(white)(aaaa)` `+` `color(white)(aaaa)` `||` `color(white)(aaaa)` `+` `color(white)(aa)` `||` `color(white)(aa)` `+`

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

`color(white)(aaaa)` `x-5` `color(white)(aaaaaa)` `-` `color(white)(aaaa)` `-` `color(white)(aaaa)` `||` `color(white)(aaaa)` `-` `color(white)(aa)` `||` `color(white)(aa)` `+`

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

Therefore,

`f(x)>0`, when `x in (-31,-1) uu (5, +oo)`