Let the original fraction = n / [n + 3]
And we have that
[n + 7] / [n + 8] = n / [n + 3] + 1/2
[ n + 7] / [n + 8] = [ 2n + n + 3] / [ 2(n + 3) ] cross-multiply
[2(n +3)] [ n + 7]= [2n + n + 3] [ n + 8]
[2(n +3)] [ n + 7]= [3n + 3] [ n + 8]
2 [ n + 3] [ n + 7 ] = 3 [n + 1] [ n + 8 ]
2 [ n ^2 + 10n + 21] = 3 [ n^2 + 9n + 8]
2n^2 + 20n + 42 = 3n^2 + 27n + 24
n ^2 + 7n - 18 = 0 factor
(n + 9) ( n - 2) = 0 and n = -9 or n = 2
So....one possibility...when n = 2
9 / 10 = 2/5 + 1/2
9/10 = 9/10
Or when n = -9
-2/-1 = -9/-6 + 1/2
2 = 3/2 + 1/2
2 = 2
So....the original fractions were 2/5 or 3/2