non linear system solving

Question

I have the following system : $$\begin{cases}2x-w-2d=0\\ 2y-w+d=0\\ w(x+y-6)=0\\ d(2x-y-4)=0\end{cases}$$ I used to solve linear equations using pivots but i don't think this will work with this system , so I need a method to follow to solve this system

Answer 1

The first two equations tell you the solution is contained in a plane in $\mathbb{R}^4$. Notice that, by the nature of the last two equations, it is convenient to express $w$ and $d$ in terms of $x$ and $y$. Once you will do this, you will find out that: $$ \Sigma_1 : \begin{cases}w = \frac{2}{3}(x + 2y)\\ d = \frac{2}{3}(x - y) \end{cases}. $$ Now, substitute in the last two equations. You will get the following system in the variables $(x,y) \in \mathbb{R}^2$: $$ \Sigma_2 : \begin{cases} (x + 2y)(x + y - 6) = 0\\ (x - y)(2x - y - 4) = 0 \end{cases} . $$ So, it corresponds to the intersection of pairs of lines in $\mathbb{R}^2$ between the four that appear in the system! So actually, you need to find the solution to solve: $$ \begin{cases} x + 2y = 0\\ x - y = 0 \end{cases} \quad \land \quad \begin{cases} x + 2y = 0\\ 2x - y - 4 = 0 \end{cases} \quad \land \quad \begin{cases} x + y - 6 = 0\\ x - y = 0 \end{cases} \quad \land \quad \begin{cases} x + y - 6 = 0\\ x - y = 0 \end{cases}. $$ If I am not mistaken, these should give the following couples: $$ (x,y) = \begin{cases} (0,0)\\ (-\frac{4}{5},\frac{8}{5})\\ (3,3)\\ (\frac{10}{3}, \frac{8}{3}) \end{cases}. $$ Now just plug every value obtained into $\Sigma_1$ to find the complete tuples of solutions $(x,y,w,d) \in \mathbb{R}^4$.