题目内容
11.方程组$\left\{\begin{array}{l}{2x+y=4}\\{x+3z=1}\\{x+y+z=7}\end{array}\right.$的解是( )| A. | $\left\{\begin{array}{l}{x=2}\\{y=2}\\{z=1}\end{array}\right.$ | B. | $\left\{\begin{array}{l}{x=2}\\{y=1}\\{z=1}\end{array}\right.$ | C. | $\left\{\begin{array}{l}{x=-2}\\{y=8}\\{z=1}\end{array}\right.$ | D. | $\left\{\begin{array}{l}{x=2}\\{y=2}\\{z=2}\end{array}\right.$ |
分析 方程组利用加减消元法求出解即可.
解答 解:$\left\{\begin{array}{l}{2x+y=4①}\\{x+3z=1②}\\{x+y+z=7③}\end{array}\right.$,
①-③得:x-z=-3④,
②-④得:4z=4,即z=1,
把z=1代入④得:x=-2,
把x=-2代入①得:y=8,
则方程组的解为$\left\{\begin{array}{l}{x=-2}\\{y=8}\\{z=1}\end{array}\right.$,
故选C.
点评 此题考查了解三元一次方程组,利用了消元的思想,消元的方法有:代入消元法与加减消元法.
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