Question

3．方程組$\left\{\begin{array}{l}{2x+3y+z=6}\\{x-y+2z=-1}\\{x+2y-z=5}\end{array}\right.$的解是：$\left\{\begin{array}{l}{x=2}\\{y=1}\\{z=-1}\end{array}\right.$．

Answer 1

分析 方程組利用加減消元法求出解即可．

解答 解：$\left\{\begin{array}{l}{2x+3y+z=6①}\\{x-y+2z=-1②}\\{x+2y-z=5③}\end{array}\right.$，
①+③得：3x+5y=11④，
②+③×2得：3x+3y=9，即x+y=3⑤，
⑤×5-④得：2x=4，即x=2，
把x=2代入⑤得：y=1，
把x=2，y=1代入③得：z=-1，
則方程組的解為$\left\{\begin{array}{l}{x=2}\\{y=1}\\{z=-1}\end{array}\right.$，
故答案為：$\left\{\begin{array}{l}{x=2}\\{y=1}\\{z=-1}\end{array}\right.$

點(diǎn)評(píng) 此題考查了解三元一次方程組，熟練掌握運(yùn)算法則是解本題的關(guān)鍵．