Solve for $x$ if $4^{\frac{x}{y} + \frac{y}{x}}$ $=$ $32$ and $\log_3(x+y)+\log_3(x-y)=1$ 

Question:
Solve for $x$ if $4^{\frac{x}{y} + \frac{y}{x}}$ $= 32$ and $\log_3(x+y)+\log_3(x-y)=1$



My attempt: With the first equation
$$4^{\frac{x}{y} + \frac{y}{x}} = 32$$
$$2^{2(\frac{x}{y} + \frac{y}{x})} = 2^5$$
$$ 2(\frac{x}{y} + \frac{y}{x}) = 5 $$
$$ \frac{x}{y} + \frac{y}{x} = \frac{5}{2} $$
$$ \frac{x^2 + y^2}{xy} = \frac{5}{2} $$
Now with the second equation
$$\log_3(x+y)+\log_3(x-y)=1$$
$$\log_3((x+y)(x-y)) = 1 $$
$$ \log_3(x^2-y^2) = 1 $$
$$ x^2-y^2 = 3$$
Now I have 2 equations:
$$ \frac{x^2 + y^2}{xy} = \frac{5}{2} $$
$$ x^2-y^2 = 3$$
Now I am stuck..
 A: 
Sorry- I am on my phone in a cafe and this napkin was the best I could do.
A: The set $$ x^2 - y^2 = 3 $$ in the plane is a hyperbola. You can draw it. What kind of set is
$$ x^2 -\frac{5}{2}xy + y^2 = 0,  $$ or
$$ 2x^2 -5xy + 2y^2 = 0?  $$
The second version factors...............
A: You have the two equations $$\frac{x^2 + y^2}{xy} = \frac{5}{2}\tag 1$$ $$x^2-y^2 = 3\tag 2$$ From $(2)$, $y=\pm \sqrt{x^2-3}$. So, plugging in $(1)$ $$\frac{x^2 + y^2}{xy} = \frac{5}{2}\implies 2(x^2+y^2)=5 x y \implies 2(2x^2-3)=\pm 5x\sqrt{x^2-3}\tag 3$$ Square both sides $$4(2x^2-3)^2=25x^2(x^2-3)$$ Expand and group terms to get $$-9 x^4+27 x^2+36=0$$ which is a quadratic in $t$ if $t=x^2$; its roots are $-1$ and $4$.
I am sure that you can take it from here.
A: $$\frac{x^2+y^2}{xy}=\frac 52$$
$$y=xt$$
$$\frac {x^2+x^2t^2}{x^2t}=\frac 52$$
$$\frac{1+t^2}{t}=\frac 52$$
$$t=2; t=\frac 12$$
$y=2x \Rightarrow x^2-4x^2=3 x^2-4x^2=3 \Rightarrow x^2=-1$
$y=\frac 12x \Rightarrow x=2y \Rightarrow 4y^2-y^2=3 \Rightarrow y=\pm 1; x =\pm 2$
