Question

5．方程组$\left\{\begin{array}{l}{2x+y=7}\\{x+2y=8}\end{array}\right.$的解为$\left\{\begin{array}{l}{x=2}\\{y=3}\end{array}\right.$．

Answer 1

分析 根据加减消元法可以求得方程组$\left\{\begin{array}{l}{2x+y=7}\\{x+2y=8}\end{array}\right.$的解，本题得以解决．

解答 解：$\left\{\begin{array}{l}{2x+y=7}&{①}\\{x+2y=8}&{②}\end{array}\right.$，
①×2-②，得3x=6，
解得x=2，
将x=2代入①，得y=3．
故原方程组的解是$\left\{\begin{array}{l}{x=2}\\{y=3}\end{array}\right.$．
故答案为：$\left\{\begin{array}{l}{x=2}\\{y=3}\end{array}\right.$．

点评 本题考查二元一次方程组的解，解题的关键是明确解二元一次方程组的方法．