<s>[INST] 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?
 [/INST] 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:
$$(x,y) \in \left\{(1, 2), (-1, -2), 
\left(\frac{5}{\sqrt{7}}, \frac{1}{\sqrt{7}}\right), 
\left(-\frac{5}{\sqrt{7}}, -\frac{1}{\sqrt{7}}\right)\right\}.$$
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