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.
 A: From the last three equations, substract the previous multiplied by $x$ to eliminate $a$:
$$b(y-x)y^2=62-24x,\\b(y-x)y=24-10x,\\b(y-x)=10-6x.$$
Now take the ratios to eliminate the common factors:
$$y=\frac{62-24x}{24-10x}=\frac{24-10x}{10-6x}$$
This is equivalent to
$$x^2-3x+1=0$$ and there are two real solutions. From $x$, compute $y$, then $b(y-x)y^3$, which leads to $ax^4+by^4$.
A: 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.
A: Define $A_i$ as follows
\begin{eqnarray*}
a+b=A_0 \\ 
ax+by=A_1 \\ 
ax^2+by^2=A_2 \\
ax^3+by^3=A_3 \\
ax^4+by^4=A_4 .
\end{eqnarray*}
Multiply the first & third equations & subtract the square of the second equation
\begin{eqnarray*}
A_0 A_2 -A_1^2 = ab(x-y)^2.
\end{eqnarray*}
Multiply the first & fourth equations & subtract the product of the second & third equations
\begin{eqnarray*}
A_0 A_3 -A_1 A_2 = ab(x+y)(x-y)^2 \\
(x+y) = \frac{A_0 A_3 -A_1 A_2}{A_0 A_2 -A_1^2}.
\end{eqnarray*}
Now multiply the first & fifth equations & subtract the square of the third equation
\begin{eqnarray*}
A_0 A_4 - A_2^2 = ab(x^2-y^2)^2 = ab(x+y)^2(x-y)^2 = \frac{(A_0 A_3 -A_1 A_2)^2}{A_0 A_2 -A_1^2}.
\end{eqnarray*}
Now plug in your values of $A_0,A_1,A_2,A_3$ and solve for $A_4$.
