Question

17．三元一次方程组$\left\{\begin{array}{l}{x+y+z=6}\\{2x-y+3z=9}\\{4x-y+2z=8}\end{array}\right.$的解是$\left\{\begin{array}{l}{x=1}\\{y=2}\\{z=3}\end{array}\right.$．

Answer 1

分析 观察题目，由于y的系数①与②③互为相反数，考虑消去y转化成二元一次方程组求解．

解答 解：$\left\{\begin{array}{l}{x+y+z=6①}\\{2x-y+3z=9②}\\{4x-y+2z=8③}\end{array}\right.$
①+②，得3x+4z=15④
②-③，-2x+z=1    ⑤
由④⑤组成方程组$\left\{\begin{array}{l}{3x+4z=15}\\{-2x+z=1}\end{array}\right.$
解这个方程组，得$\left\{\begin{array}{l}{x=1}\\{z=3}\end{array}\right.$
把x=1，z=3代入①，得y=2
所以原方程组的解为$\left\{\begin{array}{l}{x=1}\\{y=2}\\{z=3}\end{array}\right.$
故答案为：$\left\{\begin{array}{l}{x=1}\\{y=2}\\{z=3}\end{array}\right.$

点评 本题考查了三元一次方程组的解法．消去一个未知数，将三元一次方程组转化为二元一次方程组是解决本题的关键．