题目内容
3.解方程组:(1)$\left\{\begin{array}{l}x=y-3\\ y+2x=6\end{array}\right.$
(2)$\left\{\begin{array}{l}{x-2y=7}\\{1-3x=\frac{y-1}{2}}\end{array}\right.$.
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)$\left\{\begin{array}{l}{x=y-3①}\\{y+2x=6②}\end{array}\right.$,
把①代入②得:y+2y-6=6,即y=4,
把y=4代入①得:x=1,
则方程组的解为$\left\{\begin{array}{l}{x=1}\\{y=4}\end{array}\right.$;
(2)方程组整理得:$\left\{\begin{array}{l}{x-2y=7①}\\{6x+y=3②}\end{array}\right.$,
①+②×2得:13x=13,即x=1,
把x=1代入①得:y=-3,
则方程组的解为$\left\{\begin{array}{l}{x=1}\\{y=-3}\end{array}\right.$.
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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