《2025年一本计算题满分训练七年级数学全一册人教版》

(1)$\left\{\begin{array}{l} x-y= 1,\\ x+3y+z= 10,\\ x-2y-z= -2;\end{array} \right. $

答案： 解：$\left\{\begin{array}{l} x-y=1,①\\ x+3y+z=10,②\\ x-2y-z=-2,③\end{array}\right.$
②+③得：$2x+y=8$,④
①+④得：$3x=9$,解得$x=3$
把$x=3$代入①得：$3-y=1$,解得$y=2$
把$x=3$,$y=2$代入②得：$3+6+z=10$,解得$z=1$
所以方程组的解为$\left\{\begin{array}{l} x=3\\ y=2\\ z=1\end{array}\right.$

(2)$\left\{\begin{array}{l} x+y-z= 11,\\ x+z-y= 1,\\ y+z-x= 5;\end{array} \right. $

答案： $\left\{\begin{array}{l} x+y-z=11, \quad①\\ x+z-y=1, \quad②\\ y+z-x=5, \quad③\end{array}\right.$
①+②，得$2x=12$，解得$x=6$。
①+③，得$2y=16$，解得$y=8$。
②+③，得$2z=6$，解得$z=3$。
所以原方程组的解为$\left\{\begin{array}{l} x=6,\\ y=8,\\ z=3.\end{array}\right.$

(3)$\left\{\begin{array}{l} x+2y+z= 8,\\ 2x-y-z= -3,\\ 3x+y-2z= -1;\end{array} \right. $

答案： 解：$\left\{\begin{array}{l} x+2y+z=8,①\\ 2x-y-z=-3,②\\ 3x+y-2z=-1,③\end{array}\right.$
①+②，得$3x+y=5$，④
②×2-③，得$x-3y=-5$，⑤
④×3+⑤，得$10x=10$，解得$x=1$
将$x=1$代入④，得$3×1+y=5$，解得$y=2$
将$x=1$，$y=2$代入①，得$1+2×2+z=8$，解得$z=3$
所以原方程组的解为$\left\{\begin{array}{l} x=1\\ y=2\\ z=3\end{array}\right.$

(4)$\left\{\begin{array}{l} 3x-y+z= 4,\\ 2x+3y-z= 12,\\ x+y+z= 6;\end{array} \right. $

答案： 解：
$\begin{aligned}\left\{\begin{array}{l}3x - y + z = 4&(1)\\2x + 3y - z = 12&(2)\\x + y + z = 6&(3)\end{array}\right.\end{aligned}$
$(1)+(2)$得：$3x - y + z+2x + 3y - z = 4 + 12$，即$5x + 2y = 16$ $(4)$
$(2)+(3)$得：$2x + 3y - z+x + y + z = 12 + 6$，即$3x + 4y = 18$ $(5)$
$(4)×2-(5)$得：$2(5x + 2y)-(3x + 4y)=2×16 - 18$
$10x + 4y - 3x - 4y = 32 - 18$
$7x = 14$，解得$x = 2$
把$x = 2$代入$(4)$得：$5×2 + 2y = 16$，$10 + 2y = 16$，$2y = 6$，解得$y = 3$
把$x = 2$，$y = 3$代入$(3)$得：$2 + 3 + z = 6$，解得$z = 1$
所以方程组的解为$\left\{\begin{array}{l}x = 2\\y = 3\\z = 1\end{array}\right.$

(5)【一题多解】$\left\{\begin{array}{l} x+y= 5,\\ y+z= -2,\\ x+z= 3;\end{array} \right. $

答案： 解法一：加减消元法
1. 由 $x + y = 5$ 得 $x = 5 - y$；
2. 由 $y + z = -2$ 得 $z = -2 - y$；
3. 代入 $x + z = 3$：$(5 - y) + (-2 - y) = 3$，解得 $y = 0$；
4. 则 $x = 5 - 0 = 5$，$z = -2 - 0 = -2$。
解法二：三式相加法
1. 三式相加：$(x + y) + (y + z) + (x + z) = 5 + (-2) + 3$，即 $2x + 2y + 2z = 6$，得 $x + y + z = 3$；
2. 减 $y + z = -2$ 得 $x = 5$；
3. 减 $x + z = 3$ 得 $y = 0$；
4. 减 $x + y = 5$ 得 $z = -2$。
$\boxed{\left\{\begin{array}{l} x=5 \\ y=0 \\ z=-2 \end{array}\right.}$

(6)【一题多解】$\left\{\begin{array}{l} \frac {x}{2}= \frac {y}{3}= \frac {z}{4},\\ 2x-y+2z= -9;\end{array} \right. $

答案： 设$\frac{x}{2}=\frac{y}{3}=\frac{z}{4}=k$，则$x=2k$，$y=3k$，$z=4k$。
将$x=2k$，$y=3k$，$z=4k$代入$2x - y + 2z = -9$，得：
$2×2k - 3k + 2×4k = -9$
$4k - 3k + 8k = -9$
$9k = -9$
解得$k = -1$
所以$x = 2k = 2×(-1) = -2$，$y = 3k = 3×(-1) = -3$，$z = 4k = 4×(-1) = -4$
$\left\{\begin{array}{l} x=-2 \\ y=-3 \\ z=-4 \end{array}\right.$

(7)$\left\{\begin{array}{l} 3x+4y+z= 14,\\ x+5y+2z= 17,\\ 2x+2y-z= 3;\end{array} \right. $

答案： 解：$\left\{\begin{array}{l} 3x+4y+z=14,①\\ x+5y+2z=17,②\\ 2x+2y-z=3,③\end{array}\right.$
①+③，得$5x+6y=17$，④
③×2+②，得$5x+9y=23$，⑤
⑤-④，得$3y=6$，解得$y=2$
将$y=2$代入④，得$5x+12=17$，解得$x=1$
将$x=1$，$y=2$代入③，得$2+4-z=3$，解得$z=3$
所以方程组的解为$\left\{\begin{array}{l} x=1\\ y=2\\ z=3\end{array}\right.$

(8)【易错】$\left\{\begin{array}{l} 2x+3y-4z= -7,\\ \frac {x-4y}{3}= \frac {2y+3z}{2}= 2.\end{array} \right. $

答案： 由$\frac{x - 4y}{3} = 2$，得$x - 4y = 6$，即$x = 4y + 6$。
由$\frac{2y + 3z}{2} = 2$，得$2y + 3z = 4$，即$z = \frac{4 - 2y}{3}$。
将$x = 4y + 6$，$z = \frac{4 - 2y}{3}$代入$2x + 3y - 4z = -7$，
得$2(4y + 6) + 3y - 4×\frac{4 - 2y}{3} = -7$。
去分母，得$6(4y + 6) + 9y - 4(4 - 2y) = -21$。
去括号，得$24y + 36 + 9y - 16 + 8y = -21$。
合并同类项，得$41y + 20 = -21$。
移项，得$41y = -41$，解得$y = -1$。
将$y = -1$代入$x = 4y + 6$，得$x = 4×(-1) + 6 = 2$。
将$y = -1$代入$z = \frac{4 - 2y}{3}$，得$z = \frac{4 - 2×(-1)}{3} = 2$。
所以原方程组的解为$\left\{\begin{array}{l}x = 2\\y = -1\\z = 2\end{array}\right.$