$\begin{array}{|l}x^2-y^2=7 \\ x^2+xy+y^2=37\end{array}$ 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.
 A: 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 $a^2$.
A: 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
$$(4,3),\>(-4,-3), \>(-\frac{11}{\sqrt3},\frac{10}{\sqrt3}), \>(\frac{11}{\sqrt3},-\frac{10}{\sqrt3})$$
A: 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?
A: 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)
A: 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 $\cos t=\pm\sqrt{1-\sin^2t}$
A: Add the two to get:
$$y=\frac{44-2x^2}{x}$$
Substitute to the first:
$$x^2-\frac{(44-2x^2)^2}{x^2}=7 \Rightarrow \\
3x^4-169x^2+44^2=0 \Rightarrow \\
x^2=\frac{169\pm \sqrt{169^2-12\cdot 44^2}}{6}=\frac{169\pm 73}{6}=16;\frac{121}{3} \Rightarrow \\
x_{1,2}=\pm 4,x_{3,4}=\pm \frac{11}{\sqrt{3}}\\
y_{1,2}=\pm 3,y_{3,4}=\mp \frac{10}{\sqrt{3}}$$
