Find the maximum of $U (x,y) = x^\alpha y^\beta$ subject to $I = px + qy$ 
Let be $U (x,y) = x^\alpha y^\beta$. Find the maximum of the function $U(x,y)$ subject to the equality constraint $I = px + qy$.

I have tried to use the Lagrangian function to find the solution for the problem, with the equation
$$\nabla\mathscr{L}=\vec{0}$$
where $\mathscr{L}$ is the Lagrangian function and $\vec{0}=\pmatrix{0,0}$.
Using this method I have a system of $3$ equations with $3$ variables, but I can't simplify this system:
$$ax^{\alpha-1}y^\beta-p\lambda=0$$
$$\beta y^{\beta-1}x^\alpha-q\lambda=0$$
$$I=px+qx$$

The solution
The answer can be been found on the internet in any number of places. The function $U$ is a Cobb-Douglas utility function. The Cobb-Douglas function is one of the most commonly used utility functions in economics. 
The demand functions you should get are:
$$x(p,I)=\frac{\alpha I}{(\alpha+\beta)p}\qquad y(p,I)=\frac{\beta I}{(\alpha+\beta)q}$$
The  solution has a nice interpretation: the consumer spends a fraction $\frac{\alpha}{\alpha+\beta}$ of their income on good $x$ and fraction $\frac{\beta}{\alpha+\beta}$ on good $y$.
If you want to find the full working spend a minute or two searching the internet.
A simplification
Note here that you can simplify things by instead maximizing the function $V$ where $$V(x,y)=\ln U(x,y)=\alpha \ln x+\beta\ln y$$
Since $V$ is an increasing transformation of $U$ it will have the same maximizer. 
In fact you could simplify the working further by maximizing $W$ where $$W(x,y)=\frac{V(x,y)}{\alpha+\beta}=\bar{\alpha}\ln x+(1-\bar{\alpha})\ln y$$ where $\bar{\alpha}=\frac{\alpha}{\alpha+\beta}$.
