35. Suppose that $a, b, x$ and $y$ are real numbers such that
$$
a x+b y=3, \quad a x^{2}+b y^{2}=7, \quad a x^{3}+b y^{3}=16 \quad \text { and } \quad a x^{4}+b y^{4}=42 .
$$

Find the value of $a x^{5}+b y^{5}$.
35. Ans: 20 .
$a x^{3}+b y^{3}=\left(a x^{2}+b y^{2}\right)(x+y)-(a x+b y) x y$. Thus, we get $16=7(x+y)-3 x y$. Similarly, $a x^{4}+b y^{4}=\left(a x^{3}+b y^{3}\right)(x+y)-\left(a x^{2}+b y^{2}\right) x y$. So, we get $42=16(x+y)-7 x y$. Solving, we get $x+y=-14$ and $x y=-38$. Therefore,
$$
a x^{5}+b y^{5}=\left(a x^{4}+b y^{4}\right)(x+y)-\left(a x^{3}+b y^{3}\right) x y=42(-14)-16(-38)=20 .
$$