A triangle has sides x, x+4, and 3x−5. What is the possible range of x? 
A triangle has sides $x, x+4,$ $and$ $3x−5$. What is the possible range of x?

The answer to this is not:

$2x-9$ $<$ $x$ $<$ $4x-1$

Which is gotten from:
--> $(3x-5)-(x+4)<x<3x-5+x+4$
--> $3x-5-x-4 < x < 4x-1$
--> $5x-9 < x < 4x-1$
So what is the answer?
The correct answer has to be in:

__ < x < __


Hint Consider using the triangle inequality 

$$a+b>c$$

Thus you have that $$x+(x+4)>3x-5\iff 9>x\tag{1}$$
$$x+(3x-5)>x+4\iff 3x>9\iff x>3$$ Hence 

$$x\in(3, 9)\text{ or equivalently }\color{red}{3<x<9}$$

