Let the original fraction be
[tex]\dfrac{n-1}{n}[/tex]
Now let's increase both numerator and denominator by 2:
[tex]\dfrac{n+1}{n+2}[/tex]
This produces a 1/12 increase:
[tex]\dfrac{n+1}{n+2}-\dfrac{n-1}{n}=\dfrac{1}{12}[/tex]
The left hand side can be rearranged as
[tex]\dfrac{2}{n^2+2n}=\dfrac{1}{12}[/tex]
Invert both sides:
[tex]\dfrac{n^2+2n}{2}=12 \iff n^2+2n=24[/tex]
Solve the quadratic equation:
[tex]n^2+2n-24=0 \iff n=-6 \lor n=4[/tex]
So, in the first case, the original fraction is
[tex]\dfrac{n-1}{n}=\dfrac{-6-1}{-6}=\dfrac{7}{6}[/tex]
In the second case, we have
[tex]\dfrac{n-1}{n}=\dfrac{4-1}{4}=\dfrac{3}{4}[/tex]
