Question

I need help solving x^2+3xy-y^2=12 and x^2-y^2=-12

Answer 1

$substitute:x^2-y^2=-12\ to\ x^2+3xy-y^2=12\\\\3xy-12=12\\3xy=12+12\\3xy=24\ \ \ \ \ /:3\\xy=8\to x=\frac{8}{y}\\\\substitute\ to\ x^2-y^2=-12\\\\\left(\frac{8}{y}\right)^2-y^2=-12\\\\\frac{64}{y^2}=y^2-12$

$\frac{64}{y^2}=\frac{y^2-12}{1}\\\\y^2(y^2-12)=64\\\\y^4-12y^2-64=0\\\\substitute:y^2=t > 0\\\\t^2-12t-64=0$

$a=1;\ b=-12;\ c=-64\\\Delta=b^2-4ac\to\Delta=(-12)^2-4\cdot1\cdot(-64)=144+256=400\\\\t_1=\frac{-b-\sqrt\Delta}{2a};\ t_2=\frac{-b+\sqrt\Delta}{2a}\\\\\sqrt\Delta=\sqrt{400}=20\\\\t_1=\frac{12-20}{2\cdot1}=\frac{-8}{2}=-4 < 0;\ t_2=\frac{12+20}{2\cdot1}=\frac{32}{2}=16\\\\y^2=16\Rightarrow y=\pm\sqrt{16}\Rightarrow y=-4\ or\ y=4\\\\x=\frac{8}{-4}=-2\ or\ x=\frac{8}{4}=2\\\\Solution:x=-2\ and\ y=-4\ or\ x=2\ and\ y=4$