How many different pairs of positive integers (a, b) satisfy the equation 1/a+1/b=34/41 ?  Choose one.
A) 6	B) 3	C) 2	D) 1	E) 0
There is no certain way to solve 2 unknown with 1 equation. The best way is to look at the question and retrospect the most efficient way. In this question, a and b are only positive integers. So that is a big relief. Now, we can start with putting a=1,2,.. and so on till the time we are confident about one of the options.   So, we start with   a=1, we get b as -ve. out   a=2, we get b as 6. Yes (Now (a,b) = (2,6). We can directly see that (a,b)=(6,2) will also satisfy. So we have 2 possible solutions)   a=3, we get b as 3. Yes (Now we have 3 possible solutions)   a=4, we get b as fraction. Out   a=5, we get b again as some fraction. Out   a=6 already taken.   We have A,B options left. C,D,E are out.   A is 6. To have 6 as the answer, we will need one more pair like2,6and one more solution where a=b.   When a=b, we have only 1 solution = 1. So, one more solution, where a=b is not possible.   So, answer will be 1.
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How many different pairs of positive integers (a, b) satisfy the equation 1/a+1/b=34/41 ?