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第65页

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$-20$
解:$\begin{cases}x + y = 6,①\\5x + 4y = 24,②\end{cases}$
由①,得$y = 6 - x,$③
将③代入②,得$5x + 4(6 - x) = 24,$解得$x = 0.$
将$x = 0$代入③,得$y = 6,$
所以原方程组的解为$\begin{cases}x = 0\\y = 6\end{cases}$
解:$\begin{cases}m + n = 3,①\\5m - 3(m + n) = 3,②\end{cases}$
将①代入②,得$5m - 3×3 = 3,$解得$m = \frac{12}{5}.$
将$m = \frac{12}{5}$代入①,得$\frac{12}{5}+n = 3,$解得$n = \frac{3}{5},$
所以原方程组的解为$\begin{cases}m = \frac{12}{5}\\n = \frac{3}{5}\end{cases}$
解:原方程组$\begin{cases}3(x - 1) = y + 5,①\\\frac{y - 1}{3}=\frac{x}{5}+1,②\end{cases}$整理,得$\begin{cases}3x - 8 = y,③\\5y - 3x = 20,④\end{cases}$
将③代入④,得$5(3x - 8)-3x = 20,$解得$x = 5.$
将$x = 5$代入③,得$y = 7,$
所以原方程组的解为$\begin{cases}x = 5\\y = 7\end{cases}$
解:$\begin{cases}2x - y = 5,①\\4x + 3y = -10,②\end{cases}$
由①,得$y = 2x - 5,$③
把③代入②,得$4x + 3(2x - 5)= -10,$解得$x = \frac{1}{2}.$
将$x = \frac{1}{2}$代入③,得$y = -4,$
所以原方程组的解为$\begin{cases}x = \frac{1}{2}\\y = -4\end{cases}$
解:解方程组$\begin{cases}x - 5y = -2\\2x + 5y = -1\end{cases},$
将两个方程相加得:$3x=-3,$解得$x = -1,$
把$x = -1$代入$x - 5y = -2$得:$-1-5y=-2,$解得$y = \frac{1}{5}.$
原式$=x^{2}-2xy + y^{2}-(x^{2}-4y^{2})=x^{2}-2xy + y^{2}-x^{2}+4y^{2}=-2xy + 5y^{2}.$
当$x = -1,$$y = \frac{1}{5}$时,原式$=-2×(-1)×\frac{1}{5}+5×(\frac{1}{5})^{2}=\frac{3}{5}.$
解:因为关于$x,$$y$的两个方程组$\begin{cases}mx + 2ny = 4\\x + y = 1\end{cases}$与$\begin{cases}x = -2y\\nx+(m - 1)y = 3\end{cases}$有相同的解,
联立$\begin{cases}x = -2y\\x + y = 1\end{cases},$将$x = -2y$代入$x + y = 1$得:$-2y + y = 1,$解得$y = -1,$
把$y = -1$代入$x = -2y$得$x = 2.$
把$\begin{cases}x = 2\\y = -1\end{cases}$代入$\begin{cases}mx + 2ny = 4\\nx+(m - 1)y = 3\end{cases},$得$\begin{cases}2m-2n = 4\\2n-(m - 1)= 3\end{cases},$
整理得$\begin{cases}m - n = 2\\2n - m = 2\end{cases},$两式相加得$n = 4,$
把$n = 4$代入$m - n = 2$得$m = 6.$
综上,这个相同的解为$\begin{cases}x = 2\\y = -1\end{cases},$$m,$$n$的值分别为$6,$$4.$
解:$\begin{cases}3x - 2y = -13,①\\9x - 4y = -35,②\end{cases}$
由②,得$3x + 6x - 4y = -35,$
即$3x + 2(3x - 2y)= -35,$③
把①代入③,得$3x + 2×(-13)= -35,$
解得$x = -3.$
把$x = -3$代入①,得$y = 2,$
所以原方程组的解为$\begin{cases}x = -3\\y = 2\end{cases}$