Question

14．解三元一次方程组$\left\{\begin{array}{l}{x+3y+2z=10①}\\{2x+y+z=7②}\\{-x+2y+3z=12③}\end{array}\right.$应先消去未知数x，得到关于y、z的二元一次方程组$\left\{\begin{array}{l}{5y+5Z=22}\\{5y+7z=31}\end{array}\right.$，解这个二元一次方程组得$\left\{\begin{array}{l}{y=-\frac{1}{10}}\\{z=\frac{9}{2}}\end{array}\right.$，原方程组的解是$\left\{\begin{array}{l}{x=\frac{13}{10}}\\{y=-\frac{1}{10}}\\{z=\frac{9}{2}}\end{array}\right.$．

Answer 1

分析 先消去x，转化为关于y、z的二元方程组即可解决．

解答 解：$\left\{\begin{array}{l}{x+3y+2z=10}&{①}\\{2x+y+z=7}&{②}\\{-x+2y+3z=12}&{③}\end{array}\right.$
①+③得到：5y+5z=22    ④
②+2×③得到：5y+7z=31   ⑤
由④⑤得$\left\{\begin{array}{l}{5y+5z=21}\\{5y+7z=31}\end{array}\right.$解得$\left\{\begin{array}{l}{y=-\frac{1}{10}}\\{z=\frac{9}{2}}\end{array}\right.$代入②得x=$\frac{13}{10}$，$\left\{\begin{array}{l}{y=-\frac{1}{10}}\\{z=\frac{9}{2}}\end{array}\right.$
∴$\left\{\begin{array}{l}{x=\frac{13}{10}}\\{y=-\frac{1}{10}}\\{z=\frac{9}{2}}\end{array}\right.$．
故答案分别为x；y、z；$\left\{\begin{array}{l}{5y+5z=22}\\{5y+7z=31}\end{array}\right.$；$\left\{\begin{array}{l}{y=-\frac{1}{10}}\\{z=\frac{9}{2}}\end{array}\right.$；$\left\{\begin{array}{l}{x=\frac{13}{10}}\\{y=-\frac{1}{10}}\\{z=\frac{9}{2}}\end{array}\right.$．

点评 本题考查三元方程组，解题的关键是三元方程组转化为二元方程组，学会转化的数学思想，属于中考常考题型．