SOLUTION: When a and b are positive integers, a < b and ab &#8804; a + 3b, how many possible values are there for a?

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Question 248625: When a and b are positive integers, a < b and ab ≤ a + 3b, how many possible values are there for a?
Answer by jsmallt9(3758) About Me  (Show Source):
You can put this solution on YOUR website!
If we take a+%3C+b and add 3b to each side we get:
a+%2B+3b+%3C+b+%2B+3b
or
a+%2B+3b+%3C+4b

The reason we did this is that the left side above now matches the right side of the other inequality:
ab+%3C=+a+%2B+3b
Since ab+%3C=+a+%2B+3b and a+%2B+3b+%3C+4b then
ab+%3C+4b
by the transitive property.

This inequality we can solve. We can divide both sides by b (without having to be concerned about reversing the inequality because we know that b is positive):
a+%3C+4
Between this and the fact that a is also positive, the possible values for a are:
1, 2 and 3.