Question

помогите пожалуйста!!! кто поможет тому ещё 30 балов дам!!! ​

Answer 1

1) Ответ: x=1 ; y=2

$\left \{ {{y = x + 1} \atop {3x + 2y = 7}} \right. \\ \\ 3x + 2(x + 1) = 7 \\ 3x + 2x + 2 = 7 \\ 5x = 5 \\ x = 1 \\ \\ y = x + 1 \\ y = 1 + 1 = 2$

2) Ответ: x=2 ; y=0

$\left \{ {{x = 2 - y} \atop {3x + 2y = 6}} \right. \\ 3(2 - y) + 2y = 6 \\ 6 - 3y + 2y = 6 \\ y = 0 \\ \\ x = 2 - y \\ x = 2 - 0 = 2$

3) Ответ: x=-9 ; y=-7

$\left \{ {{y = x + 2} \atop {2x - 3y = 3}} \right. \\ 2x - 3(x + 2) = 3 \\ 2x - 3x - 6 = 3 \\ - x = 9 \\ x = - 9 \\ \\ y = x + 2 \\ y = - 9 + 2 = - 7$

4) Ответ: x=2 ; y=1

$\left \{ {{y = x - 1} \atop {3x - 2y = 4}} \right. \\ 3x - 2(x - 1) = 4 \\ 3x - 2x + 2 = 4 \\ x = 2 \\ \\ y = x - 1 \\ y = 2 - 1 = 1$

Answer 2

Ответ:

Заданные системы удобно решать методом подстановки, так как одна переменная в каждой системе уже выражена через другую .

$1)\ \ \left\{\begin{array}{l}y=x+1\\3x+2y=7\end{array}\right\ \ \left\{\begin{array}{l}y=x+1\\3x+2(x+1)=7\end{array}\right\ \ \left\{\begin{array}{l}y=x+1\\3x+2x+2=7\end{array}\right\\\\\\\left\{\begin{array}{l}y=x+1\\5x=5\end{array}\right\ \ \left\{\begin{array}{l}y=2\\x=1\end{array}\right\ \ \ \ Otvet:\ (\, 1\, ;\, 2\, )\ .$  

$2)\ \ \left\{\begin{array}{l}x=2-y\\3x+2y=6\end{array}\right\ \ \left\{\begin{array}{l}x=2-y\\3(2-y)+2y=6\end{array}\right\ \ \left\{\begin{array}{l}x=2-y\\6-3y+2y=6\end{array}\right\\\\\\\left\{\begin{array}{l}x=2-y\\-y=0\end{array}\right\ \ \left\{\begin{array}{l}x=2\\y=0\end{array}\right\ \ \ \ Otvet:\ (\, 2\, ;\, 0\, )\ .$

$3)\ \ \left\{\begin{array}{l}y=x+2\\2x-3y=3\end{array}\right\ \ \left\{\begin{array}{l}y=x+2\\2x-3(x+2)=3\end{array}\right\ \ \left\{\begin{array}{l}y=x+2\\2x-3x-6=3\end{array}\right\\\\\\\left\{\begin{array}{l}y=x+2\\-x=9\end{array}\right\ \ \left\{\begin{array}{l}y=-7\\x=-9\end{array}\right\ \ \ \ \ Otvet:\ (-9\ ;-7\ )\ .$

$4)\ \ \left\{\begin{array}{l}y=x-1\\3x-2y=4\end{array}\right\ \ \left\{\begin{array}{l}y=x-1\\3x-2(x-1)=4\end{array}\right\ \ \left\{\begin{array}{l}y=x-1\\3x-2x+2=4\end{array}\right\\\\\\\left\{\begin{array}{l}y=x-1\\x=2\end{array}\right\ \ \left\{\begin{array}{l}y=1\\x=2\end{array}\right\ \ \ \ \ Otvet:\ \ (\ 2\ ;\ 1\ )\ .$