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:
$$ax^{\alpha-1}y^\beta-p\lambda=0$$
$$\beta y^{\beta-1}x^\alpha-q\lambda=0$$
$$I=px+qx$$

If
$I=px+qy$,
then
$y = (I-px)/q$,
so
$x^ay^b
=x^a((I-px)/q)^b
=x^a(I-px)^b/q^b
$.
Differentiating,
we want
$\begin{array}\\
0
&=(x^a(I-px)^b)'\\
&=ax^{a-1}(I-px)^b-x^apb(I-px)^{b-1}\\
&=x^{a-1}(I-px)^{b-1}(a(I-px)-xpb)\\
&=x^{a-1}(I-px)^{b-1}(aI-apx-xpb)\\
&=x^{a-1}(I-px)^{b-1}(aI-xp(a+b))\\
\text{so}\\
x
&=\dfrac{aI}{p(a+b)}\\
\text{and}\\
y
&=(I-px)/q\\
&=(I-p\dfrac{aI}{p(a+b)})/q\\
&=(I-\dfrac{aI}{(a+b)})/q\\
&=\dfrac{(I(a+b)-aI}{q(a+b)}\\
&=\dfrac{bI}{q(a+b)}\\
\end{array}
$
