题目内容

5.解方程组:
(1)$\left\{\begin{array}{l}{\frac{x+1}{3}=2y}\\{2(x+1)-y=11}\end{array}\right.$
(2)$\left\{\begin{array}{l}{2x+2y+z=4}\\{2x+y+2z=7}\\{x+2y+2z=-6}\end{array}\right.$.

分析 (1)方程组整理后,利用加减消元法求出解即可;
(2)方程组利用加减消元法求出解即可.

解答 解:(1)方程组整理得:$\left\{\begin{array}{l}{x-6y=-1①}\\{2x-y=9②}\end{array}\right.$,
②×6-①得:11x=55,即x=5,
把x=5代入①得:y=1,
则方程组的解为$\left\{\begin{array}{l}{x=5}\\{y=1}\end{array}\right.$;
(2)$\left\{\begin{array}{l}{2x+2y+z=4①}\\{2x+y+2z=7②}\\{x+2y+2z=-6③}\end{array}\right.$,
①×2-②得:2x+3y=1④,
①×2-③得:3x+2y=14⑤,
⑤×3-④×2得:5x=40,即x=8,
把x=8代入④得:y=-5,
把x=8,y=-5代入③得:z=-2,
则方程组的解为$\left\{\begin{array}{l}{x=8}\\{y=-5}\\{z=-2}\end{array}\right.$.

点评 此题考查了解二元一次方程组,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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