题目内容
8.解方程组:(1)$\left\{\begin{array}{l}{3x+2y=47}\\{3x-2y=19}\end{array}\right.$
(2)$\left\{\begin{array}{l}{3x-2y=6}\\{2x+3y=17}\end{array}\right.$.
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.
解答 解:(1)$\left\{\begin{array}{l}{3x+2y=47①}\\{3x-2y=19②}\end{array}\right.$,
①+②得:6x=66,即x=11,
把x=11代入①得:y=7,
则方程组的解为$\left\{\begin{array}{l}{x=11}\\{y=7}\end{array}\right.$;
(2)$\left\{\begin{array}{l}{3x-2y=6①}\\{2x+3y=17②}\end{array}\right.$,
①×3+②×2得:13x=52,即x=4,
把x=4代入①得:y=3,
则方程组的解为$\left\{\begin{array}{l}{x=4}\\{y=3}\end{array}\right.$.
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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