Solve in real numbers the system of equations $x^2+2xy=5$, $y^2-3xy=-2$ Solve in real numbers the system of equations $x^2+2xy=5$, $y^2-3xy=-2$
I couldn't do this question and hence I looked at the solution which goes as follows:
if $x=0$ then $x^2+2xy=0\ne5$ hence $x\ne0$
I state that $y=lx$. Hence the system becomes:
$x^2(1+2l)=5$, $x^2(l^2-3l)=-2$ which becomes $5l^2-11l+2=0$, hence $l=2$ or $l=\frac{1}{5}$.
Hence the possible solutions are:
$(1, 2), (-1,-2), (\frac{5}{\sqrt{7}}, \frac{1}{\sqrt{7}}), (-\frac{5}{\sqrt{7}}, -\frac{1}{\sqrt{7}})$.
My question is why was I supposed to think of substituting $y=lx$? Why was this supposed to be intuitive and is there a more intuitive approach?

$\begin{align}
x^2+2xy & =  5 \\
y^2-3xy & = -2 
\end{align}$
$\begin{align}
y = & \frac{5-x^2}{2x} \\
x = & \frac{2+y^2}{3y} 
\end{align}$
$\begin{align}
y = & \frac{5-\left( \frac{2+y^2}{3y} \right)^2}{2 \left(\frac{2+y^2}{3y}\right)} \\
\end{align}$
$\begin{align}
y = & \frac{5 \cdot(9y^2)-(2+y^2)^2}{2 \cdot 3y (2+y^2)} \\
\end{align}$
$\begin{align}
{2 \cdot 3y^2 (2+y^2)} = & {5 \cdot(9y^2)-(2+y^2)^2} \\
\end{align}$
$\begin{align}
7y^4-29y^2+4=0
\end{align}$
$\begin{align}
y \in \{2,-2,\frac{1}{\sqrt{7}},\frac{-1}{\sqrt{7}}\}
\end{align}$
Plugging it into $x = \frac{2+y^2}{3y} $ we finish the problem:
$(x,y) \in \left\{(1, 2), (-1, -2), 
\left(\frac{5}{\sqrt{7}}, \frac{1}{\sqrt{7}}\right), 
\left(-\frac{5}{\sqrt{7}}, -\frac{1}{\sqrt{7}}\right)\right\}$
