Question

How do you solve `3/(x-2)<5/(x+2)` using a sign chart?

Answer 1

The solution is `x in ]-2,2 [ uu ]8, +oo [`

Explanation:

We cannot do crossing over

Let's simplify the inequality

`3/(x-2)<5/(x+2)`

`3/(x-2)-5/(x+2)<0`

`(3(x+2)-5(x-2))/((x-2)(x+2))<0`

`(3x+6-5x+10)/((x-2)(x+2))<0`

`(16-2x)/((x-2)(x+2))<0`

`(2(8-x))/((x-2)(x+2))<0`

Let `f(x)=(2(8-x))/((x-2)(x+2))`

We, now, build the sign chart

`color(white)(aaaa)` `x` `color(white)(aaaa)` `-oo` `color(white)(aaaa)` `-2` `color(white)(aaaaaaaaa)` `2` `color(white)(aaaaaa)` `8` `color(white)(aaaaaaa)` `+oo`

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

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

`color(white)(aaaa)` `8-x` `color(white)(aaaa)` `+` `color(white)(aaa)` `||` `color(white)(aaaa)` `+` `color(white)(aaa)` `||` `color(white)(aaa)` `+` `color(white)(aaaa)` `-`

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

Therefore,

`f(x)<0` when `x in ]-2,2 [ uu ]8, +oo [`