Question

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

Answer 1

分析 方程组利用加减消元法求出解即可．

解答 解：$\left\{\begin{array}{l}{x+y=-6①}\\{xy=5②}\end{array}\right.$，
由①得：y=-x-6③，
把③代入②得：x（-x-6）=5，
解得：x=-5或x=-1，
把x=-5代入③得：y=-1，把x=-1代入③得：y=-5，
则方程组的解为$\left\{\begin{array}{l}{x=-5}\\{y=-1}\end{array}\right.$或$\left\{\begin{array}{l}{x=-1}\\{y=-5}\end{array}\right.$，
故答案为：$\left\{\begin{array}{l}{x=-5}\\{y=-1}\end{array}\right.$或$\left\{\begin{array}{l}{x=-1}\\{y=-5}\end{array}\right.$

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．