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. 

This is not a nice solution.
Considering the equations $$a+b-6=0\tag 1$$ $$ax+by-10=0\tag 2$$ $$ax^2+by^2-24=0\tag 3$$ $$ax^3+by^3-62=0\tag 4$$ Let us eliminate the variables one at the time and express them as a function of $a$.
From $(1)$, $b=6-a$. Replacing in $(2)$ and assuming $a\neq6$, we have $$a x+(6-a) y-10=0 \implies y=\frac{a x-10}{a-6}\tag 5$$ Replacing $b$ and $y$ in $(3)$, we have $$\frac{a \left(-6 x^2+20 x-24\right)+44}{a-6}=0\implies x_{\pm}=\frac{5a\pm\sqrt{11} \sqrt{6 a-a^2}}{3 a}\tag 6$$ (assuming $a\neq 0$).
Let us use $x_+$ and replace in $(4)$ to end with $$\frac{22 \left(\sqrt{(6-a) a}-2 \sqrt{11} a+6 \sqrt{11}\right)}{9 \sqrt{(6-a) a}}=0 \tag 7$$ So, we need to solve $$\sqrt{(6-a) a}=2 \sqrt{11} a-6 \sqrt{11} \tag 8$$ Squaring both sides $$-a^2+6a=44 a^2-264 a+396\implies 45 a^2-270 a+396=0 \implies a_{\pm}=\frac{1}{5} \left(15\pm\sqrt{5}\right)$$ Considering $a_+$ and going backwards, we should get $$a=3+\frac{1}{\sqrt{5}}\qquad b=3-\frac{1}{\sqrt{5}}\qquad x=\frac{1}{2} \left(3+\sqrt{5}\right)\qquad y=\frac{1}{2} \left(3-\sqrt{5}\right)$$ and then the value of $(a x^n+b y^n)$ for any $n$.
Quite laborious, isn't it ? 
Thanks for the fun.
