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=(8)/(y)\\\\substitute\ to\ x^2-y^2=-12\\\\\left((8)/(y)\right)^2-y^2=-12\\\\(64)/(y^2)=y^2-12`


`(64)/(y^2)=(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=(-b-\sqrt\Delta)/(2a);\ t_2=(-b+\sqrt\Delta)/(2a)\\\\\sqrt\Delta=√(400)=20\\\\t_1=(12-20)/(2\cdot1)=(-8)/(2)=-4 < 0;\ t_2=(12+20)/(2\cdot1)=(32)/(2)=16\\\\y^2=16\Rightarrow y=\pm√(16)\Rightarrow y=-4\ or\ y=4\\\\x=(8)/(-4)=-2\ or\ x=(8)/(4)=2\\\\Solution:x=-2\ and\ y=-4\ or\ x=2\ and\ y=4`