x=2+y\\3x-y^2=6 \end{cases}\\\\3(2+y)-y^2=6\\6+3y-y^2=6\\y^2-3y=0\\y_1=0\ ;y_2=3\\x_1=2\ ;x_2=5" alt="\begin{cases} x-y=2=>x=2+y\\3x-y^2=6 \end{cases}\\\\3(2+y)-y^2=6\\6+3y-y^2=6\\y^2-3y=0\\y_1=0\ ;y_2=3\\x_1=2\ ;x_2=5" align="absmiddle" class="latex-formula">
Ответ:(2;0);(5;3)