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?

$x^2+2xy=5$, $y^2-3xy=-2$
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 $x=5y$
$$25y^2+10y^2=5\to y=\pm\frac{1}{\sqrt 7};\; \left(\frac{5}{\sqrt 7},\frac{1}{\sqrt 7}\right);\;\left(-\frac{5}{\sqrt 7},-\frac{1}{\sqrt 7}\right)$$
