题目内容
17.解方程组:(1)$\left\{\begin{array}{l}{2x-7y=8}\\{3x-8y-10=0}\end{array}\right.$
(2)$\left\{\begin{array}{l}{\frac{x+y}{2}+\frac{x-y}{3}=6}\\{4(x+y)-5(x-y)=2}\end{array}\right.$
(3)$\left\{\begin{array}{l}{\frac{x-3}{2}-3(y-1)=0}\\{2(x-3)-2(y-1)=10}\end{array}\right.$
(4)$\left\{\begin{array}{l}{a-b+c=0}\\{4a+2b+c=3}\\{9a-3b+c=28}\end{array}\right.$.
分析 (1)方程组利用加减消元法求出解即可;
(2)方程组整理后,利用加减消元法求出解即可;
(3)方程组整理后,利用加减消元法求出解即可;
(4)方程组利用加减消元法求出解即可.
解答 解:(1)方程组整理得:$\left\{\begin{array}{l}{2x-7y=8①}\\{3x-8y=10②}\end{array}\right.$,
①×8-②×7得:-5x=-6,即x=1.2,
把x=1.2代入①得:y=-0.8,
则方程组的解为$\left\{\begin{array}{l}{x=1.2}\\{y=-0.8}\end{array}\right.$;
(2)方程组整理得:$\left\{\begin{array}{l}{5x+y=6①}\\{-x+9y=2②}\end{array}\right.$,
①+②×5得:46y=16,即y=$\frac{8}{23}$;
①×9-②得:46x=52,即x=$\frac{26}{23}$,
则方程组的解为$\left\{\begin{array}{l}{x=\frac{26}{23}}\\{y=\frac{8}{23}}\end{array}\right.$;
(3)方程组整理得:$\left\{\begin{array}{l}{x-6y=-3①}\\{x-y=7②}\end{array}\right.$,
②-①得:5y=10,即y=2,
把y=2代入②得:x=9,
则方程组的解为$\left\{\begin{array}{l}{x=9}\\{y=2}\end{array}\right.$;
(4)$\left\{\begin{array}{l}{a-b+c=0①}\\{4a+2b+c=3②}\\{9a-3b+c=28③}\end{array}\right.$,
②-①得:3a+3b=3,即a+b=1④,
②-③得:-5a+5b=-25,即a-b=5⑤,
④+⑤得:2a=6,即a=3,
把a=3代入④得:b=-2,
把a=3,b=-2代入①得:c=-5,
则方程组的解为$\left\{\begin{array}{l}{a=3}\\{b=-2}\\{c=-5}\end{array}\right.$.
点评 此题考查了解二元一次方程组,熟练掌握运算法则是解本题的关键.
| A. | $\left\{\begin{array}{l}{x=-3}\\{y=2}\end{array}\right.$ | B. | $\left\{\begin{array}{l}{x=2}\\{y=-3}\end{array}\right.$ | C. | $\left\{\begin{array}{l}{x=1}\\{y=5}\end{array}\right.$ | D. | $\left\{\begin{array}{l}{x=0}\\{y=-2}\end{array}\right.$ |
| A. | ∠A+∠B+∠C+∠D=∠E | B. | ∠A+∠D+∠E=∠B+∠C | ||
| C. | ∠A+∠B+∠C+∠D-∠E=180° | D. | ∠A+∠B+∠C+∠D+∠E=360° |
| 场次 得分 班级 | 1 | 2 | 3 | 4 | 5 |
| 一班 | 85 | 88 | 77 | 75 | 85 |
| 二班 | 95 | 85 | 70 | 80 | 80 |
(2)分别求出两个班五场比赛得分的平均值;
(3)你认为哪个班级的得分较稳定?为什么?