《2025年春如金卷课时作业AB本八年级数学上册北师大版》

(1)$\left\{\begin{array}{l} x-4y+z=-3①,\\ 2x+y-z=18②,\\ x-y-z=7③\end{array}\right.$
(2)$\left\{\begin{array}{l} x+2y-z=1,\\ 3x-3y+z=2,\\ 2x+3y+z=7;\end{array}\right. $
(3)$\left\{\begin{array}{l} 2x+3y-z=11,\\ 2x+y-5z=8,\\ -2x+7y+z=19;\end{array}\right. $
(4)$\left\{\begin{array}{l} x+3y+2z=3,\\ 2x-3y-z=-2,\\ 4x+3y-3z=-2;\end{array}\right. $
(5)$\left\{\begin{array}{l} x+y+z=6,\\ x+y-z=0,\\ x-y=-1;\end{array}\right. $
(6)$\left\{\begin{array}{l} x+y+z=2,\\ x-y=3,\\ 2x-y+z=7。\end{array}\right. $

答案：
(1)解:①+②,得$3x - 3y = 15$,
即$x - y = 5$,④
②-③,得$x + 2y = 11$,⑤
由④和⑤组成一个二元一次方程组$\begin{cases}x + 2y = 11\\x - y = 5\end{cases}$
解得$x = 7$,$y = 2$,
再把$x = 7$,$y = 2$代入③,得$z = - 2$。
所以方程组的解为$\begin{cases}x = 7\\y = 2\\z = - 2\end{cases}$。
(2)$\begin{cases}x + 2y - z = 1①\\3x - 3y + z = 2②\\2x + 3y + z = 7③\end{cases}$,
①+②,得$4x - y = 3$④,
①+③,得$3x + 5y = 8$⑤,
由④和⑤组成一个二元一次方程组:
$\begin{cases}4x - y = 3\\3x + 5y = 8\end{cases}$,
解得$\begin{cases}x = 1\\y = 1\end{cases}$,
把$\begin{cases}x = 1\\y = 1\end{cases}$代入①,得$1 + 2 - z = 1$,
解得$z = 2$,
所以方程组的解是$\begin{cases}x = 1\\y = 1\\z = 2\end{cases}$。
(3)$\begin{cases}2x + 3y - z = 11①\\2x + y - 5z = 8②\\- 2x + 7y + z = 19③\end{cases}$,
①+③,得$10y = 30$,
解得$y = 3$,
②+③,得$8y - 4z = 27$④,
将$y = 3$代入④,得$z = - \frac{3}{4}$,
将$z = - \frac{3}{4}$,$y = 3$代入②,得$x = \frac{5}{8}$,
∴原方程组的解为$\begin{cases}x = \frac{5}{8}\\y = 3\\z = - \frac{3}{4}\end{cases}$。
(4)$\begin{cases}x + 3y + 2z = 3①\\2x - 3y - z = - 2②\\4x + 3y - 3z = - 2③\end{cases}$,
①+②得$3x + z = 1$④,
(②+③)÷2得$3x - 2z = - 2$⑤,
④与⑤组成方程组得$\begin{cases}3x + z = 1④\\3x - 2z = - 2⑤\end{cases}$,
解得$\begin{cases}x = 0\\z = 1\end{cases}$,
把$\begin{cases}x = 0\\z = 1\end{cases}$代入①得,$0 + 3y + 2 = 3$,
∴$y = \frac{1}{3}$,
∴方程组的解为$\begin{cases}x = 0\\y = \frac{1}{3}\\z = 1\end{cases}$。
(5)$\begin{cases}x + y + z = 6①\\x + y - z = 0②\\x - y = - 1③\end{cases}$
①+②,得:$2x + 2y = 6$,即$x + y = 3$④…
③+④,得:$2x = 2$,
∴$x = 1$…
把$x = 1$代入③,得:$1 - y = - 1$
∴$y = 2$…
把$x = 1$、$y = 2$代入②,得:$1 + 2 - z = 0$
∴$z = 3$…
所以,原方程的解是$\begin{cases}x = 1\\y = 2\\z = 3\end{cases}$…
(6)$\begin{cases}x + y + z = 2①\\x - y = 3②\\2x - y + z = 7③\end{cases}$,
③-①得,$x - 2y = 5$④,
②-④得,$y = - 2$,
将$y = - 2$代入②得,$x = 1$,
将$x = 1$,$y = - 2$代入①得,$z = 3$,
∴方程组的解为$\begin{cases}x = 1\\y = - 2\\z = 3\end{cases}$。