Solution:
We have $a x^{3} + b y^{3} = 16$, so $(a x^{3} + b y^{3})(x + y) = 16(x + y)$ and thus
$$a x^{4} + b y^{4} + x y (a x^{2} + b y^{2}) = 16(x + y)$$
It follows that
$$42 + 7 x y = 16(x + y)$$
From $a x^{2} + b y^{2} = 7$, we have $(a x^{2} + b y^{2})(x + y) = 7(x + y)$ so $a x^{3} + b y^{3} + x y (a x + b y) = 7(x + y)$. This simplifies to
$$16 + 3 x y = 7(x + y)$$
We can now solve for $x + y$ and $x y$ from (1) and (2) to find $x + y = -14$ and $x y = -38$. Thus we have $(a x^{4} + b y^{4})(x + y) = 42(x + y)$, and so $a x^{5} + b y^{5} + x y (a x^{3} + b y^{3}) = 42(x + y)$. Finally, it follows that $a x^{5} + b y^{5} = 42(x + y) - 16 x y = 20$ as desired.