$\begin{array}{|l}x^2-y^2=7 \\ x^2+xy+y^2=37\end{array}$

Question

Any suggestions how to solve: $$\begin{array}{|l}x^2-y^2=7 \\ x^2+xy+y^2=37\end{array}$$ I can simplify the system and get a homogeneous polynomial of degree $2$, but I think it must have an easier way.

Answer 1

Hint Let $$a=x+y \\ b=x-y$$

Then, the equations become $$ab=7 \\ 3a^2+b^2=4 \cdot 37$$

Thus $$3a^2+\frac{49}{a^2}=148$$

This is a quadratic in $a^2$.

Answer 2

Another possible approach consists in considering the change of variables $x = r\cos(\theta)$ and $y = r\sin(\theta)$, from whence we get \begin{align*} \begin{cases} r^{2}\cos(2\theta) = 7\\\\ r^{2}(2 + \sin(2\theta)) = 74 \end{cases} \Longleftrightarrow r^{2} = \frac{7}{\cos(2\theta)} = \frac{74}{2+\sin(2\theta)} \end{align*}

Can you take it from here?

Answer 3

Note,

$$\frac17(x^2-y^2)=\frac1{37}( x^2+xy+y^2)$$

or

$$44y^2+7xy-30x^2=0$$

which leads to $x=\frac43y$ and $x=-\frac{11}{10}y$. Plug them into $x^2-y^2=7$ to obtain the solutions

$$(4,3),\>(-4,-3), \>(-\frac{11}{\sqrt3},\frac{10}{\sqrt3}), \>(\frac{11}{\sqrt3},-\frac{10}{\sqrt3})$$

Answer 4

Observe that $xy\ne0$

WLOG $y=mx$ so that $7=x^2(1-m^2)\ \ (1)$

$$\dfrac{37}7=\dfrac{(m^2+m+1)x^2}{x^2(1-m^2)}$$

Solve the quadratic equation for $m$

Replace the value of $m$ in $(1)$

This method will work for any two homogeneous equations in $x,y$ of the same degree (which is $2$ here)

Answer 5

WLOG $x=\sqrt7\sec t,y=\sqrt7\tan t$

$$37=\dfrac{7(1+\sin t+\sin^2t)}{\cos^2\theta}$$

$$\iff37(1-\sin^2t)=7(1+\sin t+\sin^2t)$$

Solve the quadratic equation for$\sin t$

Then $\cos t=\pm\sqrt{1-\sin^2t}$

Answer 6

Add the two to get: $$y=\frac{44-2x^2}{x}$$ Substitute to the first: $$x^2-\frac{(44-2x^2)^2}{x^2}=7 \Rightarrow \\ 3x^4-169x^2+44^2=0 \Rightarrow \\ x^2=\frac{169\pm \sqrt{169^2-12\cdot 44^2}}{6}=\frac{169\pm 73}{6}=16;\frac{121}{3} \Rightarrow \\ x_{1,2}=\pm 4,x_{3,4}=\pm \frac{11}{\sqrt{3}}\\ y_{1,2}=\pm 3,y_{3,4}=\mp \frac{10}{\sqrt{3}}$$