I couldn't think of a more scientific method, but here's how I see it:
First look at what we've got and break it down into prime factors:
6 = 2*3
9 = 3*3
1. tells you that the least common multiple of x and 6 is 30.
30 = 2*3*5. We're lucky, since you can't make too many numbers out of just 2, 3 and 5, with the restriction that 5 MUST be one of x's divisors (there must be a reason for the 5 in 30 - it can't just pop out of nowhere). You can easily tell that there are only 4 possibilities for x:
a. x = 5, case when the lcm of x, 6 and 9 will be 90
b. x = 10, case when the lcm of x, 6 and 9 will be 90
c. x = 15, case when the lcm of x, 6 and 9 will be 90
d. x = 30, case when the lcm of x, 6 and 9 will be 90.
Since you get 90 all over, 1 is sufficient.
2. 45 = 3*3*5. Again, there aren't a lot of possibilities, even fewer in this case:
a. x = 15, with lcm 90
b. x = 45, with lcm 90.
Answer's 90 all over, meaning that 2 is sufficient as well.
