First of all, we can observe that
So the expression becomes
This means that the expression is defined for every 
Now, since the denominator is always positive (when it exists), the fraction can only be positive if the denominator is also positive: we must ask
Since we can't accept 3 as an answer, the actual solution set is
![(-\infty,-2] \cup [2,3) \cup (3,\infty)](https://tex.z-dn.net/?f=%28-%5Cinfty%2C-2%5D%20%5Ccup%20%5B2%2C3%29%20%5Ccup%20%283%2C%5Cinfty%29)
Answer:i dont know
Step-by-step explanation:
Looks fishy to me, the way this problem combines <, = and > symbols.
But anyway. Subtract 3x from both sides. You'll get
5 < x = 2 > 0 2 > 0 is true, but that's a different type of math problem (true or false). If you decide to substitute '1' for 2 > 0, then the "1" says "TRUE."
Then 5 < x = true, so you must find x values that are larger than 5.
That would be (5, infinity)
Please go back and ensure that you have copied down the original problem exactly as it appears.
Answer:
this cant be simplified unless you have an integer in place of "i"
