Note that $(ax^n+by^n)(x+y)=(ax^{n+1}+by^{n+1})+xy(ax^{n-1}+by^{n-1})$. Then use the above identity and solve the system, $$133(x+y)=406+49xy$$$$49(x+y)=133+7xy$$to get $x$ and $y$. Then plug those values in any two equation of the given system and solve for $a$ and $b$. Then the rest is easy arithmetic.