Question

Допоможить !!!!!!!¡!!!¡!!!

Answer 1

$\left \{ {{x-4y=6} \atop {x^2+3xy-y^2=-3}} \right. \\x=6+4y\\(6+4y)^2+3y(6+4y)-y^2+3=0\\36+48y+16y^2+18y+12y^2-y^2+3=0\\27y^2+66y+39=0\\D=4356-4*27*39=4356-4212= 144\\\sqrt{D}=12\\y_{1}=\frac{-66+12}{54}=\frac{-54}{54}=-1\\y_{2}=\frac{-66-12}{54}=-\frac{78}{54}= \frac{13}{9} \\\\x_{1} =6+4*(-1)=2\\x_{2}= 6+4*\frac{13}{9} = 6+\frac{52}{9} =\frac{54+52}{9} =\frac{106}{9} =11\frac{7}{9}$

Відповідь: (2;-1), $(11\frac{7}{9} ;\frac{13}{9} )$