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?
 A: $x^2+2xy=5$, $y^2-3xy=-2$
Multiply the first by $2$, the second by $5$ and add
$$2 x^2-11 x y+5 y^2=0$$
Divide all terms by $y^2$ and set $\frac{x}{y}=z$
$$2z^2-11z+5=0\to z=\frac{1}{2};\;z=5$$
So we have $\frac{x}{y}=\frac{1}{2}\to y=2x$ and $\frac{x}{y}=5\to x = 5y$
Plug $y=2x$ into the first equation
$$x^2+2x(2x)=5\to x=\pm 1;\; (1,2);\;(-1,-2)$$
and then $x=5y$
$$25y^2+10y^2=5\to y=\pm\frac{1}{\sqrt 7};\; \left(\frac{5}{\sqrt 7},\frac{1}{\sqrt 7}\right);\;\left(-\frac{5}{\sqrt 7},-\frac{1}{\sqrt 7}\right)$$
A: Alternative approach.

Solve in real numbers the system of equations $x^2+2xy=5$, $y^2-3xy=-2$

$3x^2 + 6xy = 15~$ and $~2y^2 - 6xy = -4.$
Adding gives $3x^2 + 2y^2 = 11.$
Then, you have that $5 - x^2 = 2xy.$
Therefore, $25 - 10x^2 + x^4 = 4x^2y^2 = (2x^2)(11 - 3x^2) = 22x^2 - 6x^4.$
Therefore, $7x^4 - 32x^2 + 25 = 0.$
Applying the quadratic equation against $x^2$ gives that $x^2 \in \{1, \frac{25}{7}\}.$
Using that $2y^2 = 11 - 3x^2$ gives that
$y^2 \in \{4, \frac{1}{7}\}.$
Therefore, there are only 8 possible solutions to the equations:
$(x,y) \in \{(\pm 1, \pm 2), (\pm \frac{5}{\sqrt{7}}, \pm \frac{1}{\sqrt{7}})\}.$
Checking each of these 8 possibilities against the original equation gives final answer of:
$$(x,y) \in \left\{(1, 2), (-1, -2), 
\left(\frac{5}{\sqrt{7}}, \frac{1}{\sqrt{7}}\right), 
\left(-\frac{5}{\sqrt{7}}, -\frac{1}{\sqrt{7}}\right)\right\}.$$
A: $\begin{align}
x^2+2xy & =  5 \\
y^2-3xy & = -2 
\end{align}$
$\begin{align}
y = & \frac{5-x^2}{2x} \\
x = & \frac{2+y^2}{3y} 
\end{align}$
$\begin{align}
y = & \frac{5-\left( \frac{2+y^2}{3y} \right)^2}{2 \left(\frac{2+y^2}{3y}\right)} \\
\end{align}$
$\begin{align}
y = & \frac{5 \cdot(9y^2)-(2+y^2)^2}{2 \cdot 3y (2+y^2)} \\
\end{align}$
$\begin{align}
{2 \cdot 3y^2 (2+y^2)} = & {5 \cdot(9y^2)-(2+y^2)^2} \\
\end{align}$
$\begin{align}
7y^4-29y^2+4=0
\end{align}$
$\begin{align}
y \in \{2,-2,\frac{1}{\sqrt{7}},\frac{-1}{\sqrt{7}}\}
\end{align}$
Plugging it into $x = \frac{2+y^2}{3y} $ we finish the problem:
$(x,y) \in \left\{(1, 2), (-1, -2), 
\left(\frac{5}{\sqrt{7}}, \frac{1}{\sqrt{7}}\right), 
\left(-\frac{5}{\sqrt{7}}, -\frac{1}{\sqrt{7}}\right)\right\}$
