Question

11．在$\left\{\begin{array}{l}{x=\frac{1}{2}}\\{y=1}\end{array}\right.$，$\left\{\begin{array}{l}{x=5}\\{y=\frac{1}{2}}\end{array}\right.$，$\left\{\begin{array}{l}{x=2}\\{y=-1}\end{array}\right.$，$\left\{\begin{array}{l}{x=1}\\{y=-\frac{1}{3}}\end{array}\right.$中，$\left\{\begin{array}{l}{x=2}\\{y=-1}\end{array}\right.$是二元一次方程组$\left\{\begin{array}{l}{2x+3y=1}\\{x-2y=4}\end{array}\right.$的解．

Answer 1

分析 ①×2+②×3得出一个关于x的方程，求出x，把x的值代入②求出y即可作出判断．

解答 解：$\left\{\begin{array}{l}{2x+3y=1①}\\{x-2y=4②}\end{array}\right.$
①×2+②×3得：7x=14，
∴x=2，
把x=2代入②得：2-2y=4，
∴y=-1．
故二元一次方程组$\left\{\begin{array}{l}{2x+3y=1}\\{x-2y=4}\end{array}\right.$的解是$\left\{\begin{array}{l}{x=2}\\{y=-1}\end{array}\right.$．
故答案为：$\left\{\begin{array}{l}{x=2}\\{y=-1}\end{array}\right.$．

点评 本题考查了解一元一次方程和二元一次方程组的应用，关键是把二元一次方程组转化成一元一次方程．