Question

5．方程组$\left\{\begin{array}{l}{x+2y=6}\\{2x-y=7}\end{array}\right.$的解是$\left\{\begin{array}{l}x=4\\ y=1\end{array}\right.$．

Answer 1

分析 方程组利用加减消元法求出解即可．

解答 解：$\left\{\begin{array}{l}{x+2y=6①}\\{2x-y=7②}\end{array}\right.$，
①+②×2得：5x=20，
解得：x=4，
把x=4代入①得：y=1，
则方程组的解为$\left\{\begin{array}{l}{x=4}\\{y=1}\end{array}\right.$，
故答案为：$\left\{\begin{array}{l}{x=4}\\{y=1}\end{array}\right.$

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．