Question

17．三元一次方程组$\left\{\begin{array}{l}{x+y=1}\\{y+z=5}\\{z+x=6}\end{array}\right.$的解是（　　）

• A． $\left\{\begin{array}{l}{x=1}\\{y=0}\\{z=4}\end{array}\right.$
• B． $\left\{\begin{array}{l}{x=1}\\{y=2}\\{z=4}\end{array}\right.$
• C． $\left\{\begin{array}{l}{x=1}\\{y=0}\\{z=5}\end{array}\right.$
• D． $\left\{\begin{array}{l}{x=4}\\{y=1}\\{z=0}\end{array}\right.$

Answer 1

分析 ①+②+③求出x+y+z=6④，④-①求出z，④-②求出x，④-③求出y．

解答 解：$\left\{\begin{array}{l}{x+y=1①}\\{y+z=5②}\\{z+x=6③}\end{array}\right.$
①+②+③得：2x+2y+2z=12，
x+y+z=6④，
④-①得：z=5，
④-②得：x=1，
④-③得：y=0，
所以原方程组的解为：$\left\{\begin{array}{l}{x=1}\\{y=0}\\{z=5}\end{array}\right.$，
故选C．

点评 本题考查了解三元一次方程组，能选择适当的方法解方程组是解此题的关键．