题目内容
12.解下列方程组(1)$\left\{\begin{array}{l}{x=2y}\\{3x-2y=8}\end{array}\right.$
(2)$\left\{\begin{array}{l}{x+\frac{1}{4}y=9}\\{\frac{1}{5}x+y=17}\end{array}\right.$.
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)$\left\{\begin{array}{l}{x=2y①}\\{3x-2y=8②}\end{array}\right.$,
把①代入②得:6y-2y=8,即y=2,
把y=2代入①得:x=4,
则方程组的解为$\left\{\begin{array}{l}{x=4}\\{y=2}\end{array}\right.$;
(2)方程组整理得:$\left\{\begin{array}{l}{4x+y=36①}\\{x+5y=85②}\end{array}\right.$,
①×5-②得:19x=85,即x=5,
把x=5代入②得:y=16,
则方程组的解为$\left\{\begin{array}{l}{x=5}\\{y=16}\end{array}\right.$.
点评 此题考查了二元一次方程组的解,方程组的解即为能使方程组中两方程都成立的未知数的值.
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