《2025年通城学典课时作业本七年级数学上册湘教版》

1. 有下列方程组：① $\left\{\begin{array}{l} x = 2,\\ y = 1,\\ z = 3\end{array} \right.$,② $\left\{\begin{array}{l} 2x = 4 - y + z,\\ 3y = 6,\\ 4z = 7\end{array} \right.$,③ $\left\{\begin{array}{l} x + y + \frac{1}{z} = 3,\\ 2x + y = 7,\\ y + 8 = 5\end{array} \right.$,④ $\left\{\begin{array}{l} xyz = 5,\\ x + y = 4,\\ y + 2z = 3\end{array} \right.$,其中,属于三元一次方程组的有(B)
A.1 个
B.2 个
C.3 个
D.4 个

答案： B

2. 运用加减消元法解方程组 $\left\{\begin{array}{l} 11x + 3z = 9,\\ 3x + 2y + z = 8,\\ 2x - 6y + 4z = 5,\end{array} \right.$ 最简便的方法是(C)
A.先消去 $x$,再解 $\left\{\begin{array}{l} 22y + 2z = 61,\\ 66y - 38z = - 37\end{array} \right.$
B.先消去 $z$,再解 $\left\{\begin{array}{l} 2x - 6y = - 15,\\ 38x + 18y = 21\end{array} \right.$
C.先消去 $y$,再解 $\left\{\begin{array}{l} 11x + 7z = 29,\\ 11x + 3z = 9\end{array} \right.$
D.三个方程相加得 $8x - 2y + 4z = 11$,再解

答案： C

3. 方程组 $\left\{\begin{array}{l} 2x + y = 3,\\ 3x - z = 7,\\ x - y + 3z = 0\end{array} \right.$ 的解为(C)
A.$\left\{\begin{array}{l} x = 2,\\ y = 1,\\ z = - 1\end{array} \right.$
B.$\left\{\begin{array}{l} x = 2,\\ y = - 1,\\ z = 1\end{array} \right.$
C.$\left\{\begin{array}{l} x = 2,\\ y = - 1,\\ z = - 1\end{array} \right.$
D.$\left\{\begin{array}{l} x = 2,\\ y = 1,\\ z = 1\end{array} \right.$

答案： C

4. 在解方程组 $\left\{\begin{array}{l} x - y + z = 0,\\ 4x + 2y + z = 3,\\ 25x + 5y + z = 60\end{array} \right.$ 时,最好先消去未知数z,得到二元一次方程组$\left\{\begin{array}{l} 3x+3y=3,\\ 21x+3y=57\end{array}\right. $(只写出一个即可)。

答案： z 答案不唯一，如$\left\{\begin{array}{l} 3x+3y=3,\\ 21x+3y=57\end{array}\right. $

5. 如图,每条边上的三个数之和都等于 16,那么 $a$,$b$,$c$ 这三个数按顺序分别为5,6,4。
]

答案： 5,6,4

6. (教材 P137 练习题变式)解下面的三元一次方程组：
(1)$\left\{\begin{array}{l} x + y - z = 11,\\ y + z - x = 5,\\ z + x - y = 1;\end{array} \right.$
(2)$\left\{\begin{array}{l} x + y + z = 12,\\ x + 2y + 5z = 22,\\ x = 4y.\end{array} \right.$

答案：
(1)$\left\{\begin{array}{l} x=6,\\ y=8,\\ z=3\end{array}\right. $
(2)$\left\{\begin{array}{l} x=8,\\ y=2,\\ z=2\end{array}\right. $

7. 已知甲、乙、丙三个数的和为 26,甲数比乙数大 1,甲数的两倍与丙数的和比乙数大 18,求这三个数。

答案： 设甲、乙、丙三个数分别为x,y,z.由题意，得$\left\{\begin{array}{l} x+y+z=26,\\ x-y=1,\\ 2x+z-y=18,\end{array}\right. $解得$\left\{\begin{array}{l} x=10,\\ y=9,\\ z=7\end{array}\right. $