题目内容
7.已知$\left\{\begin{array}{l}{x+y=5}\\{y+z=-2}\\{z+x=3}\end{array}\right.$,则x+y+z=0.分析 三式相加再两边同时除以2即可得答案.
解答 解:将三个方程相加得:2(x+y+z)=0,
∴x+y+z=0.
点评 本题虽然以三元一次方程组的形式出现,但并不需要解方程,而是考查的等式的基本性质,是很基础的一道题.熟悉等式性质并观察出三个方程中三个未知数的轮换特性是关键.
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2.下列方程组中,解为$\left\{\begin{array}{l}{x=1}\\{y=1}\\{z=2}\end{array}\right.$的是( )
| A. | $\left\{\begin{array}{l}{x+y+z=4}\\{2x+y-z=1}\\{3x+2y-4z=-3}\end{array}\right.$ | B. | $\left\{\begin{array}{l}{x-y-z=0}\\{z+y-x=1}\\{2x+y-2x=5}\end{array}\right.$ | ||
| C. | $\left\{\begin{array}{l}{x+y=4}\\{y+z=5}\\{x+z=6}\end{array}\right.$ | D. | $\left\{\begin{array}{l}{2x+3y-z=5}\\{x+y+z=4}\\{x-y+2z=2}\end{array}\right.$ |