I am trying to solve an exercise with partial derivatives. I have no issue with the derivation part, but once you do that you are supposed to solve a system of equations to find the solution. I already did a few exercises of this kind. However, I found one with a particular system of equations that I can not figure out how to solve. Multiplying one equation by something to delete one variable and solve the other one does not seem to work. I need to know the value of $\frac{\partial x}{\partial u}$ and $\frac{\partial y}{\partial u}$.
Said system of equations is:
$ \left\{ \begin{aligned} \left(2x+y\right)\frac{\partial x}{\partial u}+x\frac{\partial y}{\partial u}=-2 \\ y\frac{\partial x}{\partial u}+\left(x-2y\right)\frac{\partial y}{\partial u}=-1 \\ \end{aligned} \right. $
The answers are supposed to be: $\frac{\partial x}{\partial u}=\frac{4y-x}{2\left(x^2-2xy-y^2\right)}$ and $\frac{\partial y}{\partial u}=\frac{y-2x}{2\left(x^2-2xy-y^2\right)}$
The system of equations can be written as $$ \begin{pmatrix} 2x+y & x \\ y & x-2y \\ \end{pmatrix} \begin{pmatrix} \frac{\partial x}{\partial u} \\ \frac{\partial y}{\partial u} \end{pmatrix} = \begin{pmatrix} -2 \\ -1 \end{pmatrix} $$
Using the formula for the inverse of a $2\times 2$ matrix, we get $$ \begin{pmatrix} \frac{\partial x}{\partial u} \\ \frac{\partial y}{\partial u} \end{pmatrix} = \frac{1}{(2x+y)(x-2y)-yx} \begin{pmatrix} x-2y & -x \\ -y & 2x+y \\ \end{pmatrix} \begin{pmatrix} -2 \\ -1 \end{pmatrix} \\ = \frac{1}{2x^2-4xy-2y^2} \begin{pmatrix} (x-2y)(-2) + (-1)(-x) \\ (-y)(-2) + (2x+y)(-1) \end{pmatrix} \\ \frac{1}{2(x^2-2xy-y^2)} \begin{pmatrix} 4y - x \\ y - 2x \end{pmatrix} $$