Small confusion when solving the system of nonlinear equations $\begin{cases} 3y-2xy-y^2=0\\ 3x-x^2-2xy =0 \end{cases}$

Question

When solving the system below

\begin{cases} 3y-2xy-y^2=0 &(1)\\ 3x-x^2-2xy =0 &(2)\end{cases}

My approach is to put $y$ and$x$, respectively, as a common factor on the two equations:

$$ \begin{cases} 3y-2xy-y^2=0\\ 3x-x^2-2xy =0 \end{cases} \implies \begin{cases} y(3-2x-y)=0\\ x(3-x-2y) =0 \end{cases} $$

What I am unsure of is the next step. Should I consider only one equation (let's call it option I)? Or both of the equations (let's call it option II)?

Option I: considering only equation (1)

$y=0$:

\begin{align*} x(3-x-2\cdot 0)&=0\\ x(3-x)&=0\\ x=0 \hspace{0.5cm}&\text{ or }\hspace{0.5cm}x=3\\ (0,0)\hspace{0.5cm}&\hspace{0.5cm}(3,0) \end{align*}

$3-2x-y=0 \quad \implies \quad y= 3-2x$

\begin{align*} x[3-x-2(3-2x)]&=0\\ x(-3+3x)&=0\\ x=0 \hspace{0.5cm}&\text{ or }\hspace{0.5cm}x=1\\ (0,3)\hspace{0.5cm}&\hspace{0.5cm}(1,1) \end{align*}

Option II: considering all equation

$y=0$:

\begin{align*} x(3-x-2\cdot 0)&=0\\ x(3-x)&=0\\ x=0 \hspace{0.5cm}&\text{ or }\hspace{0.5cm}x=3\\ (0,0)\hspace{0.5cm}&\hspace{0.5cm}(3,0) \end{align*}

$3-2x-y=0 \quad \implies \quad y= 3-2x$

\begin{align*} x[3-x-2(3-2x)]&=0\\ x(-3+3x)&=0\\ x=0 \hspace{0.5cm}&\text{ or }\hspace{0.5cm}x=1\\ (0,3)\hspace{0.5cm}&\hspace{0.5cm}(1,1) \end{align*}

$x=0$:

\begin{align*} y(3-y)&=0\\ y=0 \hspace{0.5cm}&\text{ or }\hspace{0.5cm}y=3\\ (0,0)\hspace{0.5cm}&\hspace{0.5cm}(0,3) \end{align*}

$3-2y-x=0 \quad \implies \quad x= 3-2y$

\begin{align*} y[3-2(3-2y) - y]&=0\\ y(-3+3y)&=0\\ y=0 \hspace{0.5cm}&\text{ or }\hspace{0.5cm}y=1\\ (3,0)\hspace{0.5cm}&\hspace{0.5cm}(1,1) \end{align*}

If option I is the correct answer, was it really a huge coincidence that we have found the exact same solution when dealing with equations (1) and (2)?

Thank you.

Answer 1

You can take a look at the solutions of the first equation ($y = 0$ and $y = 3-2x$) and the see what happens to the second equation with these particular values of $y$. So the first approach is fine.

It is not a coincidence that you get exactly the same solutions with the second approach. You are repeating the same process as for the first equation with the second equation, and in a system both equations must be satisfied. So the first approach gives the same solutions as the second.

Answer 2

Hint:

$(1)-(2)$ gives: $$x^2 - y^2 +3y-3x = (x+y)(x-y)-3(x-y)= 0$$

Answer 3

HINT (Some traces of electrical engineering explanations may be found in the text below, please avoid if in risk of allergic reaction)

Ok so you have factored the equations. Now is where the logic comes into play. Since zero is the only number that "short-circuits" under multiplication operation, at least one factor needs to be zero in each equation at the same time. "At least one" is logical inclusive or and "at same time" is logical and.

So you get sixteen possibilities, you can put them in a truth table: You can then evaluate systematically in the truth table which combinations of equations you need to investigate. Then you solve each of those combinations one by one and collect all the solutions you find.