《2025年暑假衔接课堂七年级数学浙教版》

5. 解方程组:
(1)$\begin{cases}\dfrac{1}{x - 1} + \dfrac{2}{6y - 3} = 4, \\ \dfrac{1}{2x - 2} - \dfrac{1}{2y - 1} = 0; \end{cases}$
解：$\left\{\begin{array}{l} x=\frac {4}{3},\\ y=\frac {5}{6}.\end{array}\right. $
(2) $\begin{cases} \frac { 1 } { x } + \frac { 1 } { y } - \frac { 2 } { z } = - 4, \\ \frac { 1 } { x } - \frac { 1 } { y } + \frac { 4 } { z } = 1 1, \\ \frac { 1 } { x } + \frac { 2 } { y } = 5. \\ \end{cases}$
解：$\left\{\begin{array}{l} x=5,\\ y=\frac {5}{12},\\ z=\frac {10}{33}.\end{array}\right. $

答案： 5.
(1)$\left\{\begin{array}{l} x=\frac {4}{3},\\ y=\frac {5}{6}.\end{array}\right. $
(2)$\left\{\begin{array}{l} x=5,\\ y=\frac {5}{12},\\ z=\frac {10}{33}.\end{array}\right. $

6. 已知关于$ x , y $的二元一次方程组$ \begin{cases} a _ { 1 } x + b _ { 1 } y = c _ { 1 }, \\ a _ { 2 } x + b _ { 2 } y = c _ { 2 } \\ \end{cases}\\$　　的解是$ \begin{cases} x = 2, \\ y = 3, \\ \end{cases}\\$　　则关于$ x , y $的二元一次方程组$ \begin{cases} a _ { 1 } ( x + y ) + b _ { 1 } ( x - y ) = 2 c _ { 1 }, \\ a _ { 2 } ( x + y ) + b _ { 2 } ( x - y ) = 2 c _ { 2 } \\ \end{cases}\\$　　的解是$\left\{\begin{array}{l} x=5,\\ y=-1.\end{array}\right.$.

答案： 6.$\left\{\begin{array}{l} x=5,\\ y=-1.\end{array}\right. $

【例】解方程组:
$\begin{cases}x:y:z = 2:4:1,①\\3x - 2y + 5z = 6.②\end{cases} $
【解】由①可设$x = 2k$,$y = 4k$,$z = k$.代入②得$6k - 8k + 5k = 6$,$\therefore k = 2$.
$\therefore x = 2k = 4$,$y = 4k = 8$,$z = k = 2$.$\therefore方程组的解为\begin{cases}x = 4,\\y = 8,\\z = 2.\end{cases} $

答案： 【解析】：本题可根据已知条件中$x:y:z = 2:4:1$，设$x = 2k$，$y = 4k$，$z = k$，然后将其代入方程$3x - 2y + 5z = 6$中，求出$k$的值，最后再分别求出$x$、$y$、$z$的值。
【答案】：$\begin{cases}x = 4\\y = 8\\z = 2\end{cases}$