Question

Решите системы уравнений! Алгебра!
Под буквой a)

Answer 1

$\begin{cases}3x-y=3\\-\\3x-2y=0\end{cases}\\y=3\\3x-3=3\\3x=6\\x=2\\OTBET:(2;3)$

$\begin{cases}x+5y=7|*3\\3x+2y=-5\end{cases}\ \textless \ =\ \textgreater \ \begin{cases}3x+15y=21\\-\\3x+2y=-5\end{cases}\\13y=26\\y=2\\x+10=7\\x=-3\\OTBET:(-3;2)$

$\begin{cases}x^2-y=-2\\+\\2x+y=2\end{cases}\\x^2+2x=0\\x(x+2)=0\\x_1=0\ ;x_2=-2\\y_1=2\ ;y_2=6\\OTBET:(0;2);(-2;6)$

$\begin{cases}x-y=7=\ \textgreater \ x=7+y\\xy=-10\end{cases}\\y(7+y)=-10\\y^2+7y+10=0\\\begin{cases}y_1+y_2=-7\\y_1*y_2=10\end{cases}\\y_1=-2\ ;y_2=-5\\x_1=5\ \ \ ;x_2=2\\OTBET:(5;-2);(2;-5)$

$\begin{cases}x+y=10\\x^2-y^2=40=\ \textgreater \ (x-y)(x+y)=40\end{cases}\\10(x-y)=40\\x-y=4=\ \textgreater \ x=4+y\\4+y+y=10\\2y=6\\y=3\\x+3=10\\x=7\\OTBET:(7;3)$

$\begin{cases}x^2-3y=22\\x+y=2=\ \textgreater \ x=2-y\end{cases}\\(2-y)^2-3y=22\\4-4y+y^2-3y=22\\y^2-7y-18=0\\\begin{cases}y_1+y_2=7\\ y_1*y_2=-18\end{cases}\\y_1=9\ \ \ \ ;y_2=-2\\x_1=-7\ ;x_2=4\\OTBET:(-7;9);(4;-2)$