题目内容

6.解下列二元一次方程组:
(1)$\left\{\begin{array}{l}{66x+17y=3967}\\{25x+y=1247}\end{array}\right.$
(2)$\left\{\begin{array}{l}{80x-87y=2156}\\{22x-y=868}\end{array}\right.$
(3)$\left\{\begin{array}{l}{83x+64y=9291}\\{90x-y=3598}\end{array}\right.$.

分析 (1)方程组利用加减消元法求出解即可;
(2)方程组利用加减消元法求出解即可;
(3)方程组利用加减消元法求出解即可.

解答 解:(1)$\left\{\begin{array}{l}{66x+17y=3967①}\\{25x+y=1247②}\end{array}\right.$,
②×17-①得:359x=17232,
解得:x=48,
把x=48代入②得:y=47,
则方程组的解为$\left\{\begin{array}{l}{x=48}\\{y=47}\end{array}\right.$;
(2)$\left\{\begin{array}{l}{80x-87y=2156①}\\{22x-y=868②}\end{array}\right.$,
②×87-①得:1834x=73360,
解得:x=40,
把x=40代入②得:y=12,
则方程组的解为$\left\{\begin{array}{l}{x=40}\\{y=12}\end{array}\right.$;
(3)$\left\{\begin{array}{l}{83x+64y=9291①}\\{90x-y=3598②}\end{array}\right.$,
①+②×64得:5843x=239563,
解得:x=41,
把x=41代入②得:y=92,
则方程组的解为$\left\{\begin{array}{l}{x=41}\\{y=92}\end{array}\right.$.

点评 此题考查了解二元一次方程组,利用了消元的思想,消元的方法有:代入消元法与加减消元法.

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