《2025年一遍过八年级数学上册北师大版》

解：将②变形,得4 x+10 y+y= 5,
即2(2 x+5 y)+y= 5。③
把①代入③,得2 × 3+y= 5,解得y= -1。
把y= -1代入①,得2 x-5= 3,解得x= 4。

请你解决以下问题：

解：将②变形，得3(3x - 2y) + 2y = 19，③把①代入③，得3×5 + 2y = 19，解得y = 2。把y= 2代入①，得3x - 4 = 5，解得x = 3。所以原方程组的解为$\begin{cases}x = 3,\\y = 2。$$\end{cases}$
2(x^{2} + 4y^{2}) + xy = 36，所以xy = 36 - 2(x^{2} + 4y^{2})。③由①，得3(x^{2} + 4y^{2}) - 2xy = 47，④把③代入④，得3(x^{2} + 4y^{2}) - 2[36 - 2(x^{2} + 4y^{2})] = 47，所以x^{2} + 4y^{2} = 17，把x^{2} + 4y^{2} = 17代入③，得xy = 2。

答案： 解：
(1)将②变形，得$3(3x - 2y) + 2y = 19$，③把①代入③，得$3×5 + 2y = 19$，解得$y = 2$。把$y = 2$代入①，得$3x - 4 = 5$，解得$x = 3$。所以原方程组的解为$\begin{cases}x = 3,\\y = 2。\end{cases}$
(2)由②，得$2(x^{2} + 4y^{2}) + xy = 36$，所以$xy = 36 - 2(x^{2} + 4y^{2})$。③由①，得$3(x^{2} + 4y^{2}) - 2xy = 47$，④把③代入④，得$3(x^{2} + 4y^{2}) - 2[36 - 2(x^{2} + 4y^{2})] = 47$，所以$x^{2} + 4y^{2} = 17$，把$x^{2} + 4y^{2} = 17$代入③，得$xy = 2$。

6 [2024揭阳月考]【阅读材料】小明同学在解方程组$\left\{\begin{array}{l}\frac{2 x+3 y}{4}+\frac{2 x-3 y}{3}= 7, \\ \frac{2 x+3 y}{3}+\frac{2 x-3 y}{2}= 8\end{array}\right.$时,发现如果直接用代入消元法或加减消元法求解,运算量都比较大。聪明的他想到如果把方程组中的$(2 x+3 y)$看作一个数,把$(2 x-3 y)$看作另一个数,通过换元,可以简化运算。以下是他的解题过程：令$m= 2 x+3 y$,$n= 2 x-3 y$,则原方程组可化为$\left\{\begin{array}{l}\frac{m}{4}+\frac{n}{3}= 7, \\ \frac{m}{3}+\frac{n}{2}= 8,\end{array}\right.解得\left\{\begin{array}{l}m= 60, \\ n= -24 。\end{array}\right.$所以$\left\{\begin{array}{l}2 x+3 y= 60, \\ 2 x-3 y= -24,\end{array}\right.解得\left\{\begin{array}{l}x= 9, \\ y= 14,\end{array}\right.$所以原方程组的解为$\left\{\begin{array}{l}x= 9, \\ y= 14 。\end{array}\right.$【解决问题】请你参考小明同学的做法,解方程组：$\left\{\begin{array}{l}\frac{x+y}{3}+\frac{x-y}{5}= 2, \\ \frac{x+y}{3}-\frac{x-y}{5}= -1 。\end{array}\right.$
解：令$m = \frac{x + y}{3}$，$n = \frac{x - y}{5}$，则原方程组可化为$\begin{cases}m + n = 2,\\m - n = -1,\end{cases}$解得$\begin{cases}m = \frac{1}{2},\\n = \frac{3}{2},\end{cases}$所以$\begin{cases}\frac{x + y}{3} = \frac{1}{2},\\frac{x - y}{5} = \frac{3}{2},\end{cases}$解得$\begin{cases}x = \frac{9}{2},\\y = -3,\end{cases}$所以原方程组的解为$\begin{cases}x = \frac{9}{2},\\y = -3。\end{cases}$

答案： 解：令$m = \frac{x + y}{3}$，$n = \frac{x - y}{5}$，则原方程组可化为$\begin{cases}m + n = 2,\\m - n = -1,\end{cases}$解得$\begin{cases}m = \frac{1}{2},\\n = \frac{3}{2},\end{cases}$所以$\begin{cases}\frac{x + y}{3} = \frac{1}{2},\\frac{x - y}{5} = \frac{3}{2},\end{cases}$解得$\begin{cases}x = \frac{9}{2},\\y = -3,\end{cases}$所以原方程组的解为$\begin{cases}x = \frac{9}{2},\\y = -3。\end{cases}$