Question

18．方程组$\left\{\begin{array}{l}{x+y+z=26}\\{x-y=1}\\{2x-y+z=18}\end{array}\right.$的解是$\left\{\begin{array}{l}{x=10}\\{y=9}\\{z=7}\end{array}\right.$．

Answer 1

分析 用加减消元法将三元一次方程组转化成二元一次方程组求解即可．

解答 解：$\left\{\begin{array}{l}{x+y+z=26①}\\{x-y=1②}\\{2x-y+z=18③}\end{array}\right.$，
①-③得，-x+2y=8④，
③④组成方程组得$\left\{\begin{array}{l}{x-y=1}\\{-x+2y=8}\end{array}\right.$，
解得：$\left\{\begin{array}{l}{x=10}\\{y=9}\end{array}\right.$，
代入①得，10+9+z=26，
解得：z=7，
所以原方程组的解为$\left\{\begin{array}{l}{x=10}\\{y=9}\\{z=7}\end{array}\right.$．
故答案为：$\left\{\begin{array}{l}{x=10}\\{y=9}\\{z=7}\end{array}\right.$．

点评 此题考查解三元一次方程组，掌握逐步消元的思想是解决问题的关键．