$ A = \frac{B+C}{4} \implies 4A-B = C$ $ B = \frac{C+A}{6} \implies -A+6B=C$ Let's multiply the first equation by $ m$ and the second by $ n$, and add them: $ (4m-n)A + (6n-m) b = (m+n) C$ Let's pick $ m$ and $ n$ such that $ 4m-n = 6n-m \implies 5m = 7n$. We can use $ (m,n) = (7,5)$. Doing so, we have $ 23A+23B = 12C \implies 23(A+B) = 12C \implies \frac{C}{A+B} = \frac{23}{12}$