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Let the fraction be A/B then (A-1)/(B+2)=1/2. So B+2=2(A-1)=2A-2. Therefore B=2A-4 or 2(A-2).

There is a relationship between B and A, but no unique solution. A must be greater than 1. We can write A/B as A/(2A-4) or A/(2(A-2)). We can also write 0<A/(2A-4)<1, because we want a proper fraction, not an integer or improper fraction. Therefore A<2A-4, or A>4, assuming A and B are positive. When A=5, the fraction is 5/6 which becomes 4/8=1/2 when changed. When A=6, the fraction is 6/8. Although this cancels down to 3/4, 6/8 is still a valid fraction that fulfils the conditions. When A=7, the fraction is 7/10. So we can have many A's, and if A is odd the fraction probably won't reduce by cancelling down. So there are an infinite number of solutions for A/(2A-4) and A>4. If A=2n-1 where n is an integer ≥3, we can be sure of a reduced fraction (one that doesn't cancel).

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