Question

How do you solve `2/(x+5)>4/(x+5)+3`?

Answer 1

`x<-17/3` or `x<-5 2/3`

Explanation:

`2/(x+5)>4/(x+5)+3`

Multiply every term by `(x+5)`.

`(x+5)xx2/(x+5)>(x+5)xx4/(x+5)+3(x+5)`

`cancel((x+5))xx2/cancel(x+5)>cancel((x+5))xx4/cancel(x+5)+3x+15`

`2>4+3x+15`

`2>3x+19`

Subtract `19` from each side.

`2-19>3x`

`-17>3x`

Divide both sides by `3`.

`-17/3>(3x)/3`

`-17/3>(cancel3x)/cancel3`

`-17/3>x` or `x<-17/3`

Answer 2

The solution is `x in (-17/3, -5)`

Explanation:

You cannot do crossing over

`2/(x+5)>4/(x+5)+3`

Let's perform some simplifications

`2/(x+5)-4/(x+5)-3>0`

`(2-4-3(x+5))/(x+5)>0`

`(2-4-3x-15)/(x+5)>0`

`(-3x-17)/(x+5)>0`

Let `f(x)=(-3x-17)/(x+5)`

Let's build the sign chart

`color(white)(aaaa)` `x` `color(white)(aaaaaaa)` `-oo` `color(white)(aaaaaa)` `-17/3` `color(white)(aaaaaaa)` `-5` `color(white)(aaaaaaa)` `+oo`

`color(white)(aaaa)` `-3x-17` `color(white)(aaaaaa)` `+` `color(white)(aaaaaaa)` `-` `color(white)(aaaa)` `||` `color(white)(aaa)` `-`

`color(white)(aaaa)` `x+5` `color(white)(aaaaaaaaaa)` `-` `color(white)(aaaaaaa)` `-` `color(white)(aaaa)` `||` `color(white)(aaa)` `+`

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

Therefore,

`f(x)>0` when `x in (-17/3, -5)`

graph{2/(x+5)-4/(x+5)-3 [-24.25, 11.8, -10.45, 7.57]}