第45页 - 人教版数学补充习题七年级上下册答案

①

②

解:②-①得$： \frac {2}{3} x=3 ， $

即$ x=\frac {9}{2} ，$

将$ x=\frac {9}{2} $代入②

得$： y=-9 ，$

则方程组的解为$\begin {cases}{x=\frac {9}{2} }\\{y=-9}\end {cases}$

解:由①得$： x=-y+8③ ，$

把③代入②得$： 3(-y+8)-2 y=-1 ，$

解得$： y=5 ，$

把$ y=5 $代入 ③得$： x=-5+8=3 ，$

原方程组的解为$： \begin {cases}{x=3 }\\{y=5}\end {cases}$

解:由题意得$: \begin{cases}{2 a-b=1 \text { ① } }\\{a+2 b=8②}\end{cases}$

$\text { ① } ×2+\text { ②得: } 5 a=10 $

∴$a=2 $

$\text { 把 } a=2 \text { 代入①得: } 2 ×2-b=1 $

∴$b=3$

解$：\begin {cases}{x+y=a①}\\{5 x-3 y=-a②}\end {cases}$

$\text {①} ×3+\text { ②得: } 8 x=2 a $

∴$x=\frac {a}{4} $

∴$\frac {a}{4}+y=a $

∴$y=\frac {3\ \mathrm {a}}{4} $

∴$\begin {cases}{x=\frac {a}{4} }\\{y=\frac {3\ \mathrm {a}}{4}}\end {cases}$

解$：\begin {cases}{2 x+y=7①}\\{x+2 y=8②}\end {cases}$

$① ×2 ， $得$4 x+2 y=14③$

③ -②, 得$3 x=6 ，$∴$x=2$

将$ x=2 $代入 ①, 得$2 ×2+y=7 $

∴$y=3$

所以方程组的解是$\begin {cases}{x=2 }\\{y=3}\end {cases}$

∴$\frac {x-y}{x+y}=\frac {2-3}{2+3}=-\frac {1}{5}$

$解:\because a//b\ $

$\therefore \angle 1+\angle 2=180^{\circ}\ $

$\therefore \angle 1=180^{\circ}-\angle 2\ $

$\because 2 \angle 2-\angle 1=30^{\circ}\ $

$\therefore 2 \angle 2-\left(180^{\circ}-\angle 2\right)=30^{\circ}\ $

$2 \angle 2-180^{\circ}+\angle 2=30^{\circ}\ $

$3 \angle 2=210^{\circ}\ $

$\therefore \angle 2=70^{\circ}\ $

$\therefore \angle 3=180^{\circ}-\angle 2=180^{\circ}-70^{\circ} =110^{\circ}$

$ $

$\begin {cases}{x=2}\\{y=-1}\end {cases}$

$解:\left\{\begin{array}{r}2 x+3 y=15 m① \\5 x-3 y=-m②\end{array}\right.\ $

$\text { ① }+ \text { ② 得 } 7 x=14 m\ $

$x=2 m\ $

$将\ x=2 m\ 代入①得\ 4 m+3 y=15 m\ $

$y=\frac{11}{3} m$

$故原方程组的解为\ \left\{\begin{array}{l}x=2 m \\ y=\frac{11}{3} m\end{array}\right. .$

$ $

$解:解方程组\ \left\{\begin{array}{l}2 x+5 y=-6 \\ 3 x-5 y=16\end{array}\right. ,得\left\{\begin{array}{l}x=2 \\y=-2\end{array}\right.$

$把\ \left\{\begin{array}{l}x=2 \\ y=-2\end{array}\right.\ 代入\ \left\{\begin{array}{l}a x-b y=-4 \\ b x+a y=-8\end{array}\right.\ 可得\left\{\begin{array}{l}2 a+2 b=-4 \\2 b-2 a=-8\end{array}\right.$

$解之可得\left\{\begin{array}{l}a=1 \\b=-3\end{array}\right.$

$把\ \left\{\begin{array}{l}a=1 \\ b=-3\end{array}\right.\ 代入\ 3 {a}+2 {b} , 得3 a+2 b=-3$

$ $

$解:\because(3 x-2 y-5)^{2}+\sqrt{x-2 y+3}=0\ $

$\therefore \left\{\begin{array}{l}3 x-2 y-5=0\\ x-2 y+3=0\end{array}\right.$

$解得\left\{\begin{array}{l}x=4 \\y=\frac{7}{2}\end{array}\right.\ $

$\therefore x y-5=4 \times \frac{7}{2}-5=9,\ $

$\therefore x y-5 \text { 的平方根是 } \pm 3 .$

$ $

$解:已知\ \left\{\begin{array}{l}2(x+y)+3(x-y)=3① \\ 7(x+y)-3(x-y)=24②\end{array}\right.\ \ $

$①+ ②得:\ 9(x+y)=27 ,即\ x+y=3③\ $

$将③代入①中:\ 2 \times 3+3(x-y)=3 ,即\ x-y=-1 .$

$由\ \left\{\begin{array}{l}x+y=3 \\ x-y=-1\end{array}\right. , 解得\ \left\{\begin{array}{l}x=1 \\ y=2\end{array}\right.\ $

$ $

$解: 由原方程组, 得\ \left\{\begin{array}{l}2 x+y=3 \text {, ① } \\ 2 x-y=5 \text {. ② }\end{array}\right.\ $

$由 ① +②, 得\ 4 x=8 , 解得\ x=2 .$

$由①-②, 得\ 2 y=-2 , 解得\ y=-1 .$

$所以原方程组的解是\ \left\{\begin{array}{l}x=2, \\ y=-1 .\end{array}\right.\ $

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