电子课本网 第85页

第85页

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解:​$\begin {cases}\frac {x + 1}{3}-\frac {y + 2}{4}=0&①\\\frac {x - 3}{4}-\frac {y - 3}{3}=\frac {1}{12}&②\end {cases},$​
由①,得​$\frac {x + 1}{3}=\frac {y + 2}{4},$​
设​$\frac {x + 1}{3}=\frac {y + 2}{4}=k,$​则​$x = 3k - 1,$​​$y = 4k - 2,$​
代入②,得​$\frac {3k - 1 - 3}{4}-\frac {4k - 2 - 3}{3}=\frac {1}{12},$​
解得​$k = 1。$​
所以​$x = 3 - 1 = 2,$​​$y = 4 - 2 = 2。$​
所以原方程组的解为​$\begin {cases}x = 2\\y = 2\end {cases}$​
解:把​$\frac {x + y}{6},$​​$\frac {x - y}{10}$​分别看作一个整体,分别设为​
$m,$​​$n,$​原方程组化为​$\begin {cases}m + n = 3\\m - n = -1\end {cases},$​
两式相加得​$2m = 2,$​解得​$m = 1,$​
两式相减得​$2n = 4,$​解得​$n = 2。$​
所以​$\begin {cases}\frac {x + y}{6}=1\\\frac {x - y}{10}=2\end {cases},$​即​$\begin {cases}x + y = 6\\x - y = 20\end {cases},$​
两式相加得​$2x = 26,$​解得​$x = 13,$​
两式相减得​$2y=-14,$​解得​$y = - 7。$​
所以原方程组的解为​$\begin {cases}x = 13\\y = -7\end {cases}$​
解:设​$2^x = m,$​​$3^y = n,$​
原方程组可化为​$\begin {cases}m + n = 43\\4m - n = 37\end {cases},$​
两式相加得​$5m = 80,$​解得​$m = 16,$​
把​$m = 16$​代入​$m + n = 43,$​得​$16 + n = 43,$​
解得​$n = 27。$​
即​$\begin {cases}2^x = 16\\3^y = 27\end {cases},$​解得​$\begin {cases}x = 4\\y = 3\end {cases},$​
故原方程组的解为​$\begin {cases}x = 4\\y = 3\end {cases}$​
解:将​$\frac {2x + 3y}{2}、$​​$\frac {3x + 2y}{5}$​分别看作一个整体,
分别设为​$u,$​​$v,$​得​$\begin {cases}u = v + 2\\3u = 2v + 6\end {cases},$​
将​$u = v + 2$​代入​$3u = 2v + 6,$​得
​$3(v + 2)=2v + 6,$​
解得​$v = 0,$​
把​$v = 0$​代入​$u = v + 2,$​得​$u = 2。$​
即​$\begin {cases}\frac {2x + 3y}{2}=2\\\frac {3x + 2y}{5}=0\end {cases},$​整理得​$\begin {cases}2x + 3y = 4\\3x + 2y = 0\end {cases},$​
由​$2x + 3y = 4$​得​$x=\frac {4 - 3y}{2},$​
代入​$3x + 2y = 0$​得​$3×\frac {4 - 3y}{2}+2y = 0,$​
解得​$y = \frac {12}{5},$​
把​$y = \frac {12}{5}$​代入​$x=\frac {4 - 3y}{2}$​得
​$x=\frac {4-3×\frac {12}{5}}{2}=\frac {20 - 36}{10}=-\frac {8}{5}。$​
所以原方程组的解为​$\begin {cases}x = -\dfrac {8}{5}\\y = \dfrac {12}{5}\end {cases}$​
解:​$\begin {cases}5x + 7y = 70&①\\7x + 3y = 166&②\end {cases},$​
由​$①$​可设​$5x = 35 + 35t,$​​$7y = 35 - 35t,$​
即​$x = 7 + 7t,$​​$y = 5 - 5t,$​
代入②,得​$7(7 + 7t)+3(5 - 5t)=166,$​
解得​$t = 3,$​
所以​$x = 7 + 7×3 = 28,$​​$y = 5 - 5×3=-10。$​
所以原方程组的解为​$\begin {cases}x = 28\\y = -10\end {cases}$​