If $a, b, c, d$ are positive real numbers such that $\frac{5 a+b}{5 c+d}=\frac{6 a+b}{6 c+d}$ and $\frac{7 a+b}{7 c+d}=9$, calculate $\frac{9 a+b}{9 c+d}$.
Let $\frac{5 a+b}{5 c+d}=\frac{6 a+b}{6 c+d}=k$. Then $5 a+b=k(5 c+d)$ and $6 a+b=k(6 c+d)$. Subtracting these two equations gives $a=k c$. Now we can easily get that $b=k d$. From $\frac{7 a+b}{7 c+d}=\frac{7 k c+k d}{7 c+d}=k=9$ we get $\frac{9 a+b}{9 c+d}=9$ as well.