Solving simultaneous equation without dividing by $x$

108 Views Asked by At

I have the following equations:

$$3x^2 \dfrac{\partial x}{\partial u} + 3y^2 \dfrac{\partial y}{\partial u} = 1$$ $$y\dfrac{\partial x}{\partial u} + x\dfrac{\partial y}{\partial u} = 1$$

where I know that $x\not=y$. I am trying to solve it, but the only way I can is if I divide by $x$ or I divide by $y$, but I have no restrictiont hat these cannot be $0$, except they can't be $0$ simultaneously. Is there a method to solve without dividing by either?

2

There are 2 best solutions below

0
On BEST ANSWER

Your equations can be written in the form $$ \begin{pmatrix} 3x^2 & 3y^2 \\ y & x \end{pmatrix}\begin{pmatrix}\dfrac{\partial x}{\partial u}\\ \dfrac{\partial y}{\partial u}\end{pmatrix}=\begin{pmatrix}1\\ 1\end{pmatrix} $$ and so you only need to apply the inverse of the marix on the left-hand side.

0
On

HINT.-$$\begin{cases}3x^2A+3y^2B=1\\yA+xB=1\end{cases}\Rightarrow \begin{cases}A=\dfrac{\partial x}{\partial u}=\frac{x-3y^2}{3x^3-3y^3}\\B=\dfrac{\partial y}{\partial u}=\frac{3x^2-y}{3x^3-3y^3}\end{cases}$$