# Ideas to solve the following system of equations

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 $$\frac{\partial y}{\partial u}=\frac{y-2x}{2\left(x^2-2xy-y^2\right)}$$

• think about $v=xy$. Feb 21, 2019 at 22:29
• Can you show your work, even if you don't think it worked? This is just solving two equations in two unknowns, $\partial{x}/\partial{u}$ and $\partial{y}/\partial{u}$. Feb 21, 2019 at 22:30

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 $$\begin{pmatrix} \frac{\partial x}{\partial u} \\ \frac{\partial y}{\partial u} \end{pmatrix} = \frac{1}{(2x+y)(x-2y)-yx} \begin{pmatrix} x-2y & -x \\ -y & 2x+y \\ \end{pmatrix} \begin{pmatrix} -2 \\ -1 \end{pmatrix} \\ = \frac{1}{2x^2-4xy-2y^2} \begin{pmatrix} (x-2y)(-2) + (-1)(-x) \\ (-y)(-2) + (2x+y)(-1) \end{pmatrix} \\ \frac{1}{2(x^2-2xy-y^2)} \begin{pmatrix} 4y - x \\ y - 2x \end{pmatrix}$$