《2025年一遍过八年级数学上册北师大版》

1 解下列方程组：
(1) 一题多解 $\left\{\begin{array}{l}2 x+y= 2, \\ 8 x+3 y= 9 ;\end{array}\right.$$\begin{cases}x = \frac{3}{2},\\y = -1\end{cases}$(2) $\left\{\begin{array}{l}2 x+5 y= -10, \\ 5 x-3 y= 6 ;\end{array}\right.$$\begin{cases}x = 0,\\y = -2\end{cases}$
(3) $\left\{\begin{array}{l}x-3 y= -4, \\ \frac{x+1}{2}+y= 1 ;\end{array}\right.$$\begin{cases}x = -1,\\y = 1\end{cases}$(4) $\left\{\begin{array}{l}\frac{x-2}{2}-\frac{5-y}{3}= 1, \\ \frac{x}{0.2}-\frac{y+1}{0.3}= 5 。\end{array}\right.$$\begin{cases}x = \frac{9}{2},\\y = \frac{17}{4}\end{cases}$

答案： 解：
(1)$\begin{cases}2x + y = 2,①\\8x + 3y = 9,②\end{cases}$通解(加减消元法) ② - ①×3，得$2x = 3$，解得$x = \frac{3}{2}$。把$x = \frac{3}{2}$代入①，得$2×\frac{3}{2} + y = 2$，解得$y = -1$。所以原方程组的解为$\begin{cases}x = \frac{3}{2},\\y = -1。\end{cases}$另解(代入消元法) 由①，得$y = 2 - 2x$，③把③代入②，得$8x + 3(2 - 2x) = 9$，解得$x = \frac{3}{2}$。把$x = \frac{3}{2}$代入③，得$y = -1$。所以原方程组的解为$\begin{cases}x = \frac{3}{2},\\y = -1。\end{cases}$
(2)$\begin{cases}2x + 5y = -10,①\\5x - 3y = 6,②\end{cases}$①×3 + ②×5，得$31x = 0$，解得$x = 0$。把$x = 0$代入②，得$-3y = 6$，解得$y = -2$。所以原方程组的解为$\begin{cases}x = 0,\\y = -2。\end{cases}$
(3)将原方程组整理，得$\begin{cases}x - 3y = -4,①\\x + 2y = 1,②\end{cases}$② - ①，得$5y = 5$，解得$y = 1$。把$y = 1$代入①，得$x - 3 = -4$，解得$x = -1$。所以原方程组的解为$\begin{cases}x = -1,\\y = 1。\end{cases}$
(4)将原方程组化简，得$\begin{cases}3x + 2y = 22,①\\3x - 2y = 5,②\end{cases}$① + ②，得$6x = 27$，解得$x = \frac{9}{2}$。把$x = \frac{9}{2}$代入②，得$3×\frac{9}{2} - 2y = 5$，解得$y = \frac{17}{4}$。所以原方程组的解为$\begin{cases}x = \frac{9}{2},\\y = \frac{17}{4}。\end{cases}$

解方程组：$\left\{\begin{array}{l}13 x+14 y= 40, \\ 14 x+13 y= 41 。\end{array}\right.$
解：$\begin{cases}13x + 14y = 40,①\\14x + 13y = 41,②\end{cases}$① + ②，得$27x + 27y = 81$，即$x + y = 3$，③① - ②，得$-x + y = -1$，④③ + ④，得$2y = 2$，解得$y = $1。把$y = $1代入③，得$x = $2。所以原方程组的解为$\begin{cases}x = 2,\\y = 1。\end{cases}$

答案： 解：$\begin{cases}13x + 14y = 40,①\\14x + 13y = 41,②\end{cases}$① + ②，得$27x + 27y = 81$，即$x + y = 3$，③① - ②，得$-x + y = -1$，④③ + ④，得$2y = 2$，解得$y = 1$。把$y = 1$代入③，得$x = 2$。所以原方程组的解为$\begin{cases}x = 2,\\y = 1。\end{cases}$

解方程组：$\left\{\begin{array}{l}20 x+19 y= 17, \\ 17 x+16 y= 14 。\end{array}\right.$
解：$\begin{cases}20x + 19y = 17,①\\17x + 16y = 14,②\end{cases}$① - ②，得3x + 3y = 3，所以x + y = 1，③③×16，得16x + 16y = 16，④② - ④，得x = -2，把x = -2代入③，得y = 3。所以方程组的解为$\begin{cases}x = -2,\\y = 3。\end{cases}$

答案： 解：$\begin{cases}20x + 19y = 17,①\\17x + 16y = 14,②\end{cases}$① - ②，得$3x + 3y = 3$，所以$x + y = 1$，③③×16，得$16x + 16y = 16$，④② - ④，得$x = -2$，把$x = -2$代入③，得$y = 3$。所以方程组的解为$\begin{cases}x = -2,\\y = 3。\end{cases}$

4 解下列方程组：
(1) $\left\{\begin{array}{l}\frac{2}{3}(2 x+y)= 4, \\ \frac{3}{4} x+\frac{5}{6}(2 x+y)= 8 ;\end{array}\right.$
解：$\begin{cases}\frac{2}{3}(2x + y) = 4,①\\frac{3}{4}x + \frac{5}{6}(2x + y) = 8,②\end{cases}$由①，得$2x + y =$6，③把③代入②，得$\frac{3}{4}x + \frac{5}{6}×6 = 8$，解得$x =$4。把$x = 4$代入③，得$8 + y = 6$，解得$y =$-2。所以原方程组的解为$\begin{cases}x =$4,\\y =$-2。\end{cases}$
(2) $\left\{\begin{array}{l}3 x+2 y-2= 0, \\ \frac{3 x+2 y+1}{5}-2 x= -\frac{2}{5} 。\end{array}\right.$
解：$\begin{cases}3x + 2y - 2 = 0,①\\frac{3x + 2y + 1}{5} - 2x = -\frac{2}{5},②\end{cases}$由①，得$3x + 2y =$2，③把③代入②，得$\frac{3}{5} - 2x = -\frac{2}{5}$，解得$x =$$\frac{1}{2}$。把$x = \frac{1}{2}$代入③，得$\frac{3}{2} + 2y = 2$，解得$y =$$\frac{1}{4}$。所以原方程组的解为$\begin{cases}x =$$\frac{1}{2}$,\\y =$$\frac{1}{4}$。\end{cases}$

答案： 解：
(1)$\begin{cases}\frac{2}{3}(2x + y) = 4,①\\\frac{3}{4}x + \frac{5}{6}(2x + y) = 8,②\end{cases}$由①，得$2x + y = 6$，③把③代入②，得$\frac{3}{4}x + \frac{5}{6}×6 = 8$，解得$x = 4$。把$x = 4$代入③，得$8 + y = 6$，解得$y = -2$。所以原方程组的解为$\begin{cases}x = 4,\\y = -2。\end{cases}$
(2)$\begin{cases}3x + 2y - 2 = 0,①\\\frac{3x + 2y + 1}{5} - 2x = -\frac{2}{5},②\end{cases}$由①，得$3x + 2y = 2$，③把③代入②，得$\frac{3}{5} - 2x = -\frac{2}{5}$，解得$x = \frac{1}{2}$。把$x = \frac{1}{2}$代入③，得$\frac{3}{2} + 2y = 2$，解得$y = \frac{1}{4}$。所以原方程组的解为$\begin{cases}x = \frac{1}{2},\\y = \frac{1}{4}。\end{cases}$