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2025年文轩图书假期生活指导暑七年级数学
注：目前有些书本章节名称可能整理的还不是很完善，但都是按照顺序排列的，请同学们按照顺序仔细查找。练习册 2025年文轩图书假期生活指导暑七年级数学 答案主要是用来给同学们做完题方便对答案用的，请勿直接抄袭。
1. 二元一次方程组$\left\{\begin{array}{l} 3x+2y= 12,\\ 2x-y= 1\end{array} \right. $的解为
$\left\{\begin{array}{l} x=2\\ y=3\end{array}\right.$
.
答案：
解：$\left\{\begin{array}{l} 3x+2y=12,①\\ 2x-y=1,②\end{array}\right.$
由②得$y=2x-1$，③
将③代入①，得$3x+2(2x-1)=12,$
解得$x=2,$
将$x=2$代入③，得$y=3,$
所以方程组的解为$\left\{\begin{array}{l} x=2\\ y=3\end{array}\right.$
由②得$y=2x-1$，③
将③代入①，得$3x+2(2x-1)=12,$
解得$x=2,$
将$x=2$代入③，得$y=3,$
所以方程组的解为$\left\{\begin{array}{l} x=2\\ y=3\end{array}\right.$
2. 解方程组:
$\left\{\begin{array}{l} x-2y= 1,①\\ 3x+4y= 23.②\end{array} \right. $
$\left\{\begin{array}{l} x-2y= 1,①\\ 3x+4y= 23.②\end{array} \right. $
答案：
解:①×2+②得:5x=25,解得:x=5,将x=5代入①得:5-2y=1,解得:y=2,所以原方程组的解是$\left\{\begin{array}{l} x=5,\\ y=2.\end{array}\right.$
3. 已知x,y满足方程组$\left\{\begin{array}{l} x+y= 3,\\ 2x+y= 5,\end{array} \right. 则3x+2y$=
8
.
答案：
解：
$\left\{\begin{array}{l} x+y=3, \quad(1)\\ 2x+y=5, \quad(2)\end{array}\right.$
$(2)-(1)$，得$x=2$。
将$x=2$代入$(1)$，得$2+y=3$，解得$y=1$。
则$3x+2y=3×2 + 2×1=6 + 2=8$。
8
$\left\{\begin{array}{l} x+y=3, \quad(1)\\ 2x+y=5, \quad(2)\end{array}\right.$
$(2)-(1)$，得$x=2$。
将$x=2$代入$(1)$，得$2+y=3$，解得$y=1$。
则$3x+2y=3×2 + 2×1=6 + 2=8$。
8
4. 阅读材料:解方程组
$\left\{\begin{array}{l} x-y-1= 0,①\\ 4(x-y)-y= 5.②\end{array} \right. $时,可由①得$x-y= $1③,然后再将③代入②,得$4×1-y= 5$,求得$y= -1$,再把$y= -1$代入③,解得$x= 0.从而求得原方程组的解为\left\{\begin{array}{l} x= 0,\\ y= -1.\end{array} \right. $这种方法被称为“整体代入法”.
请用上述方法解方程组:
$\left\{\begin{array}{l} 3x+2y-2= 0,\\ \frac {3x+2y+1}{5}-2x= -\frac {2}{5}.\end{array} \right. $
$\left\{\begin{array}{l} x-y-1= 0,①\\ 4(x-y)-y= 5.②\end{array} \right. $时,可由①得$x-y= $1③,然后再将③代入②,得$4×1-y= 5$,求得$y= -1$,再把$y= -1$代入③,解得$x= 0.从而求得原方程组的解为\left\{\begin{array}{l} x= 0,\\ y= -1.\end{array} \right. $这种方法被称为“整体代入法”.
请用上述方法解方程组:
$\left\{\begin{array}{l} 3x+2y-2= 0,\\ \frac {3x+2y+1}{5}-2x= -\frac {2}{5}.\end{array} \right. $
答案：
解:$\left\{\begin{array}{l} 3x+2y-2=0,①\\ \frac {3x+2y+1}{5}-2x=-\frac {2}{5}②\end{array}\right.$
由①得$3x+2y=2$③,
把③代入②,得$\frac {2+1}{5}-2x=-\frac {2}{5}$,
解得$x=\frac {1}{2}$,
把$x=\frac {1}{2}$代入③,得$\frac {3}{2}+2y=2$,
解得$y=\frac {1}{4}$,
$\therefore$原方程组的解为$\left\{\begin{array}{l} x=\frac {1}{2},\\ y=\frac {1}{4}.\end{array}\right.$
由①得$3x+2y=2$③,
把③代入②,得$\frac {2+1}{5}-2x=-\frac {2}{5}$,
解得$x=\frac {1}{2}$,
把$x=\frac {1}{2}$代入③,得$\frac {3}{2}+2y=2$,
解得$y=\frac {1}{4}$,
$\therefore$原方程组的解为$\left\{\begin{array}{l} x=\frac {1}{2},\\ y=\frac {1}{4}.\end{array}\right.$
(1) 举一反三:运用上述方法解方程组$\left\{\begin{array}{l} (\frac {a}{3}-1)+2(\frac {b}{5}+2)= 4,\\ 2(\frac {a}{3}-1)+(\frac {b}{5}+2)= 5.\end{array} \right. $
解:设$\frac{a}{3}-1=x$，$\frac{b}{5}+2=y$，原方程组可变为$\left\{\begin{array}{l}x+2y=4\\2x+y=5\end{array}\right.$
解方程组，得$\left\{\begin{array}{l}x=2\\y=1\end{array}\right.$
即$\left\{\begin{array}{l}\frac{a}{3}-1=2\\frac{b}{5}+2=1\end{array}\right.$
解得$\left\{\begin{array}{l}a=9\\b=-5\end{array}\right.$
(2) 能力运用:已知关于x,y的方程组$\left\{\begin{array}{l} a_{1}x+b_{1}y= c_{1},\\ a_{2}x+b_{2}y= c_{2}\end{array} \right. 的解为\left\{\begin{array}{l} x= 5,\\ y= 3,\end{array} \right. $则关于m,n的方程组$\left\{\begin{array}{l} a_{1}(m+3)+b_{1}(n-2)= c_{1},\\ a_{2}(m+3)+b_{2}(n-2)= c_{2}\end{array} \right. $的解是____;
(3) 拓展提高:若方程组$\left\{\begin{array}{l} 3a_{1}x+2b_{1}y= 5c_{1},\\ 3a_{2}x+2b_{2}y= 5c_{2}\end{array} \right. 的解是\left\{\begin{array}{l} x= 3,\\ y= 4,\end{array} \right. 则方程组\left\{\begin{array}{l} a_{1}x+b_{1}y= c_{1},\\ a_{2}x+b_{2}y= c_{2}\end{array} \right. $的解是____.
解:设$\frac{a}{3}-1=x$，$\frac{b}{5}+2=y$，原方程组可变为$\left\{\begin{array}{l}x+2y=4\\2x+y=5\end{array}\right.$
解方程组，得$\left\{\begin{array}{l}x=2\\y=1\end{array}\right.$
即$\left\{\begin{array}{l}\frac{a}{3}-1=2\\frac{b}{5}+2=1\end{array}\right.$
解得$\left\{\begin{array}{l}a=9\\b=-5\end{array}\right.$
(2) 能力运用:已知关于x,y的方程组$\left\{\begin{array}{l} a_{1}x+b_{1}y= c_{1},\\ a_{2}x+b_{2}y= c_{2}\end{array} \right. 的解为\left\{\begin{array}{l} x= 5,\\ y= 3,\end{array} \right. $则关于m,n的方程组$\left\{\begin{array}{l} a_{1}(m+3)+b_{1}(n-2)= c_{1},\\ a_{2}(m+3)+b_{2}(n-2)= c_{2}\end{array} \right. $的解是____;
$\left\{\begin{array}{l}m=2\\n=5\end{array}\right.$
(3) 拓展提高:若方程组$\left\{\begin{array}{l} 3a_{1}x+2b_{1}y= 5c_{1},\\ 3a_{2}x+2b_{2}y= 5c_{2}\end{array} \right. 的解是\left\{\begin{array}{l} x= 3,\\ y= 4,\end{array} \right. 则方程组\left\{\begin{array}{l} a_{1}x+b_{1}y= c_{1},\\ a_{2}x+b_{2}y= c_{2}\end{array} \right. $的解是____.
$\left\{\begin{array}{l}x=\frac{9}{5}\\y=\frac{8}{5}\end{array}\right.$
答案：
(1)解:设$\frac{a}{3}-1=x$，$\frac{b}{5}+2=y$，原方程组可变为$\left\{\begin{array}{l}x+2y=4\\2x+y=5\end{array}\right.$
解方程组，得$\left\{\begin{array}{l}x=2\\y=1\end{array}\right.$
即$\left\{\begin{array}{l}\frac{a}{3}-1=2\\frac{b}{5}+2=1\end{array}\right.$
解得$\left\{\begin{array}{l}a=9\\b=-5\end{array}\right.$
(2)$\left\{\begin{array}{l}m=2\\n=5\end{array}\right.$
(3)$\left\{\begin{array}{l}x=\frac{9}{5}\\y=\frac{8}{5}\end{array}\right.$
(1)解:设$\frac{a}{3}-1=x$，$\frac{b}{5}+2=y$，原方程组可变为$\left\{\begin{array}{l}x+2y=4\\2x+y=5\end{array}\right.$
解方程组，得$\left\{\begin{array}{l}x=2\\y=1\end{array}\right.$
即$\left\{\begin{array}{l}\frac{a}{3}-1=2\\frac{b}{5}+2=1\end{array}\right.$
解得$\left\{\begin{array}{l}a=9\\b=-5\end{array}\right.$
(2)$\left\{\begin{array}{l}m=2\\n=5\end{array}\right.$
(3)$\left\{\begin{array}{l}x=\frac{9}{5}\\y=\frac{8}{5}\end{array}\right.$
6. 解方程组:$\left\{\begin{array}{l} x:y:z= 4:7:8,\\ x+y+2z= 54.\end{array} \right. $
答案：
解：设$x = 4k$，$y = 7k$，$z = 8k$（$k$为常数）。
将其代入$x + y + 2z = 54$，得：
$4k + 7k + 2×8k = 54$
$4k + 7k + 16k = 54$
$27k = 54$
$k = 2$
则$x = 4×2 = 8$，$y = 7×2 = 14$，$z = 8×2 = 16$。
所以方程组的解为$\left\{\begin{array}{l} x=8\\ y=14\\ z=16\end{array}\right.$
将其代入$x + y + 2z = 54$，得：
$4k + 7k + 2×8k = 54$
$4k + 7k + 16k = 54$
$27k = 54$
$k = 2$
则$x = 4×2 = 8$，$y = 7×2 = 14$，$z = 8×2 = 16$。
所以方程组的解为$\left\{\begin{array}{l} x=8\\ y=14\\ z=16\end{array}\right.$
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