Determine $ax^4 + by^4$ for system of equations I found the following recreational problem without further specification for $a,b$. 
Let $x,y$ be real numbers s.t.
$a + b = 6$, 
$ax + by = 10$, 
$ax^2 + by^2 = 24$,
$ax^3 + by^3 = 62$.
Determine $ax^4 + by^4$. 
I am new to problem solving exercises like this and therefore appreciate diverse approaches to this problem as well as comments on how to tackle those types of exercises. 

I prefer to add another answer for the most general case.
Consider the four equations which I shall write $$ax^{i-1}+b y^{i-1}=c_i \qquad (i=1,2,3,4)$$ Manipulating them, we can show that $x y=\lambda$ and $a b=\mu$ using $$\lambda=\frac{c_3^2-c_2 c_4}{c_2^2-c_1 c_3}$$
$$\mu=-\frac{\left(c_2^2-c_1 c_3\right)^3}{(4 c_4 c_2^3-3 c_3)^2 c_2^2-6 c_1 c_2
   c_3 c_4+c_1 \left(4 c_3^3+c_1 c_4^2\right)}$$ This makes $$a_{\pm}=\frac{1}{2} \left(c_1\pm\sqrt{c_1^2-4 \mu }\right)$$ Using $a_+$ , this gives $$b=c_1-a\qquad x=\frac{c_2+\sqrt{c_2^2-4 \lambda  \mu }}{2 a}\qquad y=\frac \lambda x$$ Using the given values for the $c_i$'s, this leads to $$\lambda=1\qquad \mu=\frac{44}{5}$$ $$a=3+\frac{1}{\sqrt{5}}\qquad b=3-\frac{1}{\sqrt{5}}\qquad x=\frac{1}{2} \left(3+\sqrt{5}\right)\qquad y=\frac{1}{2} \left(3-\sqrt{5}\right)$$
Now, play with the numbers of your choice.
