Determining x given some probabilities

I need to determine x given the following conditions.

A and B are two events of Ω.

P(not A) = 3x

P(B) = 1/2

P(A or B) = 9x

P(A and B) = 3x

Here's what I thought of (but I know is wrong):

P(A or B) = 9x
P(A) + P(B) = 9x
1 - P(not A) + 1/2 = 9x
1 - 3x + 1/2 = 9x
x = 1/8


The answer is 0.1, but I can't get to it. I can't do the above because P(A or B) = P(A) + P(B) can only be done if P(A and B) = 0, and we are not told that here.

Any ideas? Thank you in advance.

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With $P(1-A)$ you mean $P(\text{not } A)$, I presume? –  Lord_Farin Oct 20 '12 at 12:36
Indeed, will edit it. –  user996056 Oct 20 '12 at 12:40
$P(A\cup B) = 9x$ doesn't imply $P(A) + P(B) = 9x$ when $A\cap B\ne\emptyset$. –  FrenzY DT. Oct 20 '12 at 12:51
Really @FrenzYDT.? Damn, I was pretty sure about that. –  user996056 Oct 20 '12 at 13:03
You need to use that $A = (A \cup B)\setminus B \cup (A \cap B)$. Since $B \subset A \cup B$ and $(A \cup B)\setminus B \cap (A \cap B) = \emptyset$, you have $$P(A) = P(A \cup B) - P(B) + P(A \cap B)$$ and get $$1 - 3x = 9x - \frac{1}{2} + 3x$$ which yields $x = \frac{1}{10}$.