Example 17 (Problem from the 8th American Invitational Mathematics Examination) If real numbers $a, b, x, y$ satisfy $a x+b y=3, a x^{2}+b y^{2}=$ $7, a x^{3}+b y^{3}=16, a x^{4}+b y^{4}=42$, find the value of $a x^{5}+b y^{5}$.
Solve: Since $a x^{3}+b y^{3}=16$, then $\left(a x^{3}+b y^{3}\right)(x+y)=16(x+y)$.
Therefore, $\left(a x^{4}+b y^{4}\right)+x y\left(a x^{2}+b y^{2}\right)=16(x+y)$,
which means $42+7 x y=16(x+y)$.
Also, from $a x^{2}+b y^{2}=7$, we have $\left(a x^{2}+b y^{2}\right)(x+y)=7(x+y)$,
which means $\left(a x^{3}+b y^{3}\right)+x y(a x+b y)=7(x+y)$, or $16+3 x y=7(x+y)$.
From (1) and (2), we get $\left\{\begin{array}{l}16(x+y)-7 x y=42, \\ 7(x+y)-3 x y=16,\end{array}\right.$
Solving these, we get $x+y=-14, x y=-38$.
Also, from $a x^{4}+b y^{4}=42$, we have $\left(a x^{4}+b y^{4}\right)(x+y)=42(x+y)$,
thus $\left(a x^{5}+b y^{5}\right)+x y\left(a x^{3}+b y^{3}\right)=42(x+y)$.
Therefore, $a x^{5}+b y^{5}=42(x+y)-16 x y=42 \cdot(-14)-16 \cdot(-38)=20$.