How many positive integer values of $x$ are solutions to the inequality $10 < -x + 13$?
To solve this inequality, I want to isolate $x$ on one side of the inequality sign. I can do this by subtracting 13 from both sides, which gives me $-3 < -x$. Then, I can multiply both sides by -1, but I have to remember to flip the inequality sign, since multiplying by a negative number reverses the order of the inequality. This gives me $3 > x$, or equivalently, $x < 3$. Now, I have to find how many positive integers are less than 3. The only positive integers that are less than 3 are 1 and 2, so there are only two possible values of $x$ that satisfy the inequality. Therefore, the answer is 2. 