System of equations involving 4 variables 
If $$a + b = 6$$
$$ax + by = 10$$
$$ax^2 + by^2 = 24$$
$$ax^3 + by^3 = 62$$
then $$ax^4 + by^4 = ?$$

I got $$a(x-1) + b(y-1) = 4$$ $$ax(x-1) + by(y-1) = 14$$ $$ax^2(x-1) + by^2(y-1) = 38$$ by subtracting the given equations and also got $$a(x-1)^2 + b(y-1)^2 = 10$$ $$ax^2(x-1)^2 + by^2(y-1)^2 = 24$$ by further subtracting the three equations.
I don't think I'm going anywhere with this process and so I'm not sure how to approach this. Any help is appreciated.

Use succssive eliminations.
From $(1)$, $b=6-a$. PLug in $(2)$ to get $y=\frac{a x-10}{a-6}$. Plug in $(3)$ to get $a=\frac{22}{3 x^2-10 x+12}$. Plug in $(4)$ and obtain
$$22\frac{x^2-3x+1}{3 x-5}=0 \implies x= ???$$
Now, go backward.
