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 $$42+7 x y=16(x+y) \tag{1}$$ From $a x^{2}+b y^{2}=7$, we have $(a x^{2}+b y^{2})(x+y)=7(x+y)$ so $a x^{3}+b y^{3}+x y(a x^{2}+b y^{2})=7(x+y)$. This simplifies to $$16+3 x y=7(x+y) \tag{2}$$ 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.