# System of equations: $x=xe_3-2ye_4$, $y=e_3y+e_4x$, $(\forall x,y \in R)$

I have system of two equations:
$$x=xe_3-2ye_4$$ $$y=e_3y+e_4x$$
I need to find $e_3$ and $e_4$ but they have to be independent from $x$ and $y$.
In other words these two equations need to have unique solutions $e_3$ and $e_4$ for every $x$,$y$ from $R$, where $R$ is set of real numbers $(\forall x,y \in R)$.

I tried to solve the system by substitution and by method of opposite coefficients (elimination method) but failed to succeed.

By substitution:
$$x=\frac{-2e_4y}{1-e_3}$$ $$y(1-e_3)=-\frac{2e_4^2y}{1-e_3}$$ $$1-2e_3+e_3^2+2e_4^2=0$$
And I don't know what to do now. Thee same thing happens when I try to solve it

By method of opposite coefficients: $$x(e_3-1)-2e_4y=0$$ $$xe_4+(e_3-1)y=0$$ I multiply first equation with $(e_3-1)$ and second one with $2e_4$ and add the equations to lose $y$, then I get:
$$2xe_4^2+x(e_3-1)^2=0$$ $$2e_4^2+e_3^2-2e_3+1=0$$

With both methods I get the same result and don't know what to do next.
Did I do something wrong or am I using the wrong methods? If thats the case what methods should I use to solve this system?

Solution, if someone needs it:

$e_3=1$, $e_4=0$

$$2e_4^2+e_3^2-2e_3+1=2e_4^2+(e_3-1)^2=0$$ $$\implies e_4=0 \quad ; (e_3-1)=0$$ Since a square is always non-negative, sum of two squares is zero only if both are individually zero.

• Simple...but for some reason I couldn't think of it...thank you :) Commented Oct 26, 2017 at 21:59

$x(1-e_{3})+2ye_{4}=0$

$y(1-e_{3})-xe_{4}=0$

It implies:

$xy(1-e_{3})+2y^{2}e_{4}=0$

$xy(1-e_{3})-x^{2}e_{4}=0$

Then, $e^{4}(x^{2}+2y^{2})=0$

In the other hand, we also have

$(x^{2}+2y^{2})(e^{3}-1)=0$