Easy-Peasy Algebra

Dividing both the constraints, we have $\frac{x}{y}-\frac{y}{x}=\frac{15}{4}$ let $a=\frac{x}{y}$ , then we have $a-\frac{1}{a}=\frac{15}{4}$ and hence we have the quadratic equation, $4a^2-15a-4=0$ and solving we have $a=4,\frac{-1}{4}$ .

Short Note: There are two solutions to this problem. Not one.

$\begin{aligned} \dfrac{x}{y} & = 80\\ y & = \dfrac{80}{x}\\ \text{Substituting this value in the first equation}\\ x^2 - {\left(\dfrac{80}{x}\right)}^2 & = 300\\ x^2 - \dfrac{6400}{x^2} & = 300\\ x^4 - 6400 & = 300x^2\\ x^4 - 300x^2 - 6400 & = 0\\ \text{Apply quadratic formula}\\ x & = \pm17.88855\\ \implies y & = \pm 4.47\\ \therefore \dfrac{x}{y} & \approx \boxed{4} \end{aligned}$