Question

3x/x+2 > 5

solve it by case 1. if x>0 and case 2. if x<0

Answer 1

is that \[\frac{3x}{x+2}>5\]?

Answer 2

yes

Answer 3

\[\frac{3x}{x+2}+\frac{-5x-10}{x+2} > 0\]\[\frac{-2x-10}{x+2} > 0\]\[\frac{2x+10}{x+2} < 0\]

So either 1. 2x+10<0 and x+2<0 or 2. 2x+10>0 and x+2>0

For 1.

2x+10<0 and x+2<0
2x<-10 x<-2
x<-5

So x<-2 is for the first part

NOw 2.

2x+10>0 and x+2>0
2x>-10 x>-2
x>-5

So x>-5 is for the second one

Therefore the answer is -5<x<-2

Answer 4

Sorry it took a while. Let me know if there are any questions :D

Answer 5

Like Zed, I'd prefer to solve it by case 1. x+2 > 0 and case 2. x+2 < 0. That way I can multiply both sides by x+2 yet still know whether or not to flip the inequality (yes for case 2 only). For 3x vs 5x+10 remove 3x from both sides, subtract 10, divide by 2, and check by case.