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

Question

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?

Answer 1

$x^2+2xy=5$, $y^2-3xy=-2$

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

Answer 2

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:

$$(x,y) \in \left\{(1, 2), (-1, -2), \left(\frac{5}{\sqrt{7}}, \frac{1}{\sqrt{7}}\right), \left(-\frac{5}{\sqrt{7}}, -\frac{1}{\sqrt{7}}\right)\right\}.$$

Answer 3

$\begin{align} x^2+2xy & = 5 \\ y^2-3xy & = -2 \end{align}$

$\begin{align} y = & \frac{5-x^2}{2x} \\ x = & \frac{2+y^2}{3y} \end{align}$

$\begin{align} y = & \frac{5-\left( \frac{2+y^2}{3y} \right)^2}{2 \left(\frac{2+y^2}{3y}\right)} \\ \end{align}$

$\begin{align} y = & \frac{5 \cdot(9y^2)-(2+y^2)^2}{2 \cdot 3y (2+y^2)} \\ \end{align}$

$\begin{align} {2 \cdot 3y^2 (2+y^2)} = & {5 \cdot(9y^2)-(2+y^2)^2} \\ \end{align}$

$\begin{align} 7y^4-29y^2+4=0 \end{align}$

$\begin{align} y \in \{2,-2,\frac{1}{\sqrt{7}},\frac{-1}{\sqrt{7}}\} \end{align}$

Plugging it into $x = \frac{2+y^2}{3y} $ we finish the problem:

$(x,y) \in \left\{(1, 2), (-1, -2), \left(\frac{5}{\sqrt{7}}, \frac{1}{\sqrt{7}}\right), \left(-\frac{5}{\sqrt{7}}, -\frac{1}{\sqrt{7}}\right)\right\}$