1. (教材例题变式)下列各方程组中，不是三元一次方程组的是()

A.$\begin{cases}x = 1,\\x + y = 3,\\x + y + z = 7\end{cases}$
B.$\begin{cases}x + y = 2,\\y + z = 3,\\z + x = 1\end{cases}$
C.$\begin{cases}x + 2y - z = 1,\\2x - y + z = 3,\\3x + y - 2z = 5\end{cases}$
D.$\begin{cases}x + y - z = 5,\\xyz = 1,\\x - 2y = 3\end{cases}$

1. D 解析：D选项的方程组中，$xyz = 1$是三次方程，不是一次方程.

2. 解方程组$\begin{cases}3x - y + z = 4,\\2x - z = 12,\\x + y + z = 6,\end{cases}$若要使运算简便，消元的方法应( )

A.先消$x$
B.先消$y$
C.先消$z$
D.先消哪个未知数都一样

2. B 解析：由于第二个方程只有两个未知数，且不含$y$，
∴先用第一、三两个方程消去$y$可使运算简便.

3. 已知$\frac{a}{3} = \frac{b}{5} = \frac{c}{7}$，且$3a + 2b - 4c = 9$，则$a + b + c$的值为.

3. $-15$ 解析：设$\frac{a}{3}=\frac{b}{5}=\frac{c}{7}=k$，则$a = 3k$，$b = 5k$，$c = 7k$，代入$3a + 2b - 4c = 9$，得$9k + 10k - 28k = 9$，解得$k = -1$，
∴$a = -3$，$b = -5$，$c = -7$，
∴$a + b + c = -3 + (-5) + (-7) = -15$.

4. 已知三元一次方程组$\begin{cases}x + y = 10,\\y + z = 20,\\z + x = 40,\end{cases}$则$x + y + z$的值为 ______ .

4. $35$ 解析：三个方程相加，得$2x + 2y + 2z = 70$，
∴$x + y + z = 35$.

5. 若$(2x - 4)^2 + (x + y)^2 + |4z - y| = 0$，则$x + y + z$的值为.

5. $-\frac{1}{2}$ 解析：由题意，得$\begin{cases}2x - 4 = 0,\\x + y = 0,\\4z - y = 0,\end{cases}$ 解得$\begin{cases}x = 2,\\y = -2,\\z = -\frac{1}{2},\end{cases}$
∴$x + y + z = 2 + (-2) + (-\frac{1}{2}) = -\frac{1}{2}$.

6. 解下列方程组：
(1) $\begin{cases}x + y + z = 22,\\3x + y = 47,\\x - 4z = 2;\end{cases}$
(2) $\begin{cases}x + 2y + 3z = 4,\\y + z = 1,\\3x + 2z = 3.\end{cases}$

6. (1)$\begin{cases}x + y + z = 22①,\\3x + y = 47②,\\x - 4z = 2③.\end{cases}$ ② - ①，得$2x - z = 25$④，由③④组成方程组$\begin{cases}x - 4z = 2,\\2x - z = 25,\end{cases}$ 解得$\begin{cases}x = 14,\\z = 3.\end{cases}$ 把$x = 14$代入②，得$3×14 + y = 47$，解得$y = 5$.
∴原方程组的解为$\begin{cases}x = 14,\\y = 5,\\z = 3.\end{cases}$
(2)$\begin{cases}x + 2y + 3z = 4①,\\y + z = 1②,\\3x + 2z = 3③,\end{cases}$ ① - ②×2，得$x + z = 2$④，④×2 - ③，得$-x = 1$，解得$x = -1$，把$x = -1$代入④，得$-1 + z = 2$，解得$z = 3$，把$z = 3$代入②，得$y + 3 = 1$，解得$y = -2$，原方程组的解为$\begin{cases}x = -1,\\y = -2,\\z = 3.\end{cases}$

7. 在等式$y = ax^2 + bx + c$中，当$x = 1$时，$y = - 2$；当$x = - 1$时，$y = 20$；当$x = \frac{3}{2}$与$x = \frac{1}{3}$时，$y$的值相等.求$a$、$b$、$c$的值.

7. 根据题意，得$\begin{cases}a + b + c = -2①,\\a - b + c = 20②,\\11a + 6b = 0③,\end{cases}$ ① - ②，得$2b = -22$，解得$b = -11$，将$b = -11$代入③，得$11a + 6×(-11) = 0$，解得$a = 6$，将$a = 6$，$b = -11$代入①，得$6 + (-11) + c = -2$，解得$c = 3$.

根据题意，得
$\begin{cases}a + b + c = -2 \quad ①, \\a - b + c = 20 \quad ②, \\(\frac{3}{2})^2a + \frac{3}{2}b + c = (\frac{1}{3})^2a + \frac{1}{3}b + c \quad ③\end{cases}$
化简③，得$\frac{9}{4}a + \frac{3}{2}b = \frac{1}{9}a + \frac{1}{3}b$，两边同乘36，得$81a + 54b = 4a + 12b$，整理得$77a + 42b = 0$，即$11a + 6b = 0$ ④。
① - ②，得$2b = -22$，解得$b = -11$。
将$b = -11$代入④，得$11a + 6×(-11) = 0$，解得$a = 6$。
将$a = 6$，$b = -11$代入①，得$6 + (-11) + c = -2$，解得$c = 3$。
所以$a = 6$，$b = -11$，$c = 3$。

8. 已知三个二元一次方程$3x - y = 7$，$2x + 3y = 1$，$y = kx - 9$有公共解，则$k$的值是()

A.$3$
B.$-\frac{16}{3}$
C.$- 2$
D.$4$

8. D 解析：$\begin{cases}3x - y = 7①\\2x + 3y = 1②\end{cases}$，①×3，得$9x - 3y = 21$③，② + ③，得$11x = 22$，解得$x = 2$，把$x = 2$代入①，得$3×2 - y = 7$，解得$y = -1$，将$\begin{cases}x = 2,\\y = -1\end{cases}$代入$y = kx - 9$，得$2k - 9 = -1$，解得$k = 4$.