how to solve this system of equations with 4 variables? I was looking through the UKMT mentoring samples (sample no. 9) and found this question:
solve the following system of equations:
$a+b=18$ 
$au+bv = 110$ 
$au^2+bv^2 = 690$ 
$au^3+bv^3 = 4430$
I tried dividing the equations by each other but just went in circles. I also tried subtracting the equations from each other and factorising out different parts but I didn't get anywhere.
 A: 
Long Hint:
After ruling out the possibility of a solution with $a=0$, we can multiply the third equation by $a$ and the fourth equation by $a^{2}$ to get
$$\begin{align}
(au)^{2}+abv^{2}&=690a\\
(au)^{3}+a^{2}bv^{3}&=4430a^{2}.\\
\end{align}$$
Using the second equation to isolate $au=110-bv$, the above pair becomes
$$\begin{align}
(110-bv)^{2}+abv^{2}&=690a\\
(110-bv)^{3}+a^{2}bv^{3}&=4430a^{2}.\\
\end{align}$$
Next, use the first equation to isolate $a=18-b$, we're left with the following pair of equations in two variables
$$\begin{align}
(110-bv)^{2}+(18-b)bv^{2}&=690(18-b)\\
(110-bv)^{3}+(18-b)^{2}bv^{3}&=4430(18-b)^{2}.\\
\end{align}$$
We can solve the quadratic equation for $u$ in terms of $b$ or vice versa, and plug the result into the cubic to produce a single equation in a single variable.
This has the potential to become an extremely messy calculation, but in this case the end result is simpler than you might expect! Can you take it from there? :)

A: $$a+b=18\tag 1$$
$$au+bv = 110\tag 2$$
$$au^2+bv^2 = 690\tag 3$$
$$au^3+bv^3 = 4430\tag 4$$
Use successive eliminations
$$(1)\quad \implies \quad b=18-a$$ Plug in the next
$$(2)\quad \implies \quad v=\frac{a u-110}{a-18}$$ Plug in the next
$$(3)\quad \implies \quad a=-\frac{650}{9 u^2-110 u+300}$$ Plug in the next and ... finish to get a quadratic in $u$. Then, go back to $(a,v,b)$ for the final result.
