Originally Posted by

**jam2011** can i use this:

since we have (3x-6)/(3-x)<0 then

case 1: suppose x is positive, x>0

**No! you should be supposing that the entire denominator is positive.**

Probably the examples your instructor used had "x only" in the denominator

(3x-6)/(3-x)>0 **you've inadvertently switched the inequality, by imagining x>3**

**(3x-6)/(3-x)<0**

3x-6>0 by multiplying 3-x both side

**true if x>3, as the denominator is negative and we reverse the inequality**

3x>6 by adding 6 both side

x>2

**However, x must also be >3, so x>3 means x>2 also**

Solution is x>3

Case 2 : when x is negative then x<0

**You need x<3 so that the denominator is positive**

(3x-6)/(3-x)<0

3x-6>0 by multiplying 3-x both side.... **no**

**You don't switch the inequality this time because you are not multiplying by a negative number.**

3x>6 by adding 6 both side

x>2

**3x-6<0**

**3x<6**

**x<2**

i use this pattern since this my instructor discussion? or i cannot apply it this time