(2*x^2-3*x*y-8*y^2)/(x^2+5*x*y-4*y^2)=-3/5
2*x^2-3*x*y-8*y^2)/(x^2+5*x*y-4*y^2)=-3/5
\(\frac{2x^2-3xy-8y^2}{x^2+5xy-4y^2}=\frac{-3}{5}\\ 5(2x^2-3xy-8y^2)=-3(x^2+5xy-4y^2)\\ 10x^2-15xy-40y^2=-3x^2-15xy+12y^2\\ 13x^2-52y^2=0\\ x^2-4y^2=0\\ (x-2y)(x+2y)=0\\ x=\pm2y\qquad or\\ y=\pm\frac{x}{2}\)
