《2025年暑假衔接课堂七年级数学浙教版》

【例】解三元一次方程组$\left\{\begin{array}{l}x + 2y - z = 1,\textcircled{1}\\2x - y + z = - 2,\textcircled{2}\\x = y - z.\textcircled{3}\end{array} \right.$
【解】将③分别代入①②,消去$x$,得$\left\{\begin{array}{l}3y - 2z = 1,\\y - z = - 2.\end{array} \right.$
解这个二元一次方程组,得$\left\{\begin{array}{l}y = 5,\\z = 7.\end{array} \right.$代入③,得$x = -2$.
$\therefore所以原方程组的解是\left\{\begin{array}{l}x = -2,\\y = 5,\\z = 7.\end{array} \right.$

答案： 【解析】：本题采用代入消元法来解三元一次方程组。首先将方程③$x = y - z$代入方程①$x + 2y - z = 1$，得到$(y - z)+2y - z = 1$，化简可得$3y - 2z = 1$；再将方程③代入方程②$2x - y + z = - 2$，得到$2(y - z)-y + z = - 2$，化简可得$y - z = - 2$，这样就得到了一个关于$y$和$z$的二元一次方程组$\left\{\begin{array}{l}3y - 2z = 1\\y - z = - 2\end{array}\right.$。然后求解这个二元一次方程组，由$y - z = - 2$可得$y = z - 2$，将其代入$3y - 2z = 1$中，得到$3(z - 2)-2z = 1$，即$3z - 6 - 2z = 1$，解得$z = 7$。把$z = 7$代入$y = z - 2$，可得$y = 7 - 2 = 5$。最后把$y = 5$，$z = 7$代入方程③$x = y - z$，可得$x = 5 - 7 = - 2$。
【答案】：$\left\{\begin{array}{l}x = - 2\\y = 5\\z = 7\end{array}\right.$

1. 观察方程组$\left\{\begin{array}{l}3x - y + 2z = 3,\\2x + y - 4z = 11,\\7x + y - 5z = 1\end{array} \right.$的系数特点,若要使求解简便,消元的方法应选取(B)
A.先消去$x$
B.先消去$y$
C.先消去$z$
D.以上说法都不对

答案： B

2. 由方程组$\left\{\begin{array}{l}2x + y = 7,\\2y + z = 8,\\2z + x = 9,\end{array} \right.可以得到x + y + z$的值等于(A)
A.8
B.9
C.10
D.11

答案： A

3. 写一个解为$\left\{\begin{array}{l}x = 1,\\y = 2,\\z = 3\end{array} \right.$的三元一次方程组：$\left\{\begin{array}{l}x + y + z = 6\\x + y - z = 0\\2x + y - z = 1\end{array}\right.$（答案不唯一）.

答案： $\left\{\begin{array}{l}x + y + z = 6\\x + y - z = 0\\2x + y - z = 1\end{array}\right.$（答案不唯一）