题目内容
3.解方程组:(1)$\left\{\begin{array}{l}{x-2y=1}\\{3x-5y=8}\end{array}\right.$(用代入消元法解方程组)
(2)$\left\{\begin{array}{l}{3x-2(x+2y)=3}\\{11x+4(x+2y)=45}\end{array}\right.$.
分析 (1)方程组利用代入消元法求出解即可;
(2)方程组整理后,利用加减消元法求出解即可.
解答 解:(1)$\left\{\begin{array}{l}{x-2y=1①}\\{3x-5y=8②}\end{array}\right.$,
由①得:x=2y+1③,
把③代入②得:6y+3-5y=8,即y=5,
把y=5代入①得:x=11,
则方程组的解为$\left\{\begin{array}{l}{x=11}\\{y=5}\end{array}\right.$;
(2)方程组整理得:$\left\{\begin{array}{l}{x-4y=3①}\\{15x+8y=45②}\end{array}\right.$,
①×2+②得:17x=51,即x=3,
把x=3代入①得:y=0,
则方程组的解为$\left\{\begin{array}{l}{x=3}\\{y=0}\end{array}\right.$.
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.
练习册系列答案
相关题目