Question

7．三元一次方程組$\left\{\begin{array}{l}{2x-3y+2z=5}\\{x-2y+3z=-6}\\{3x-y+z=3}\end{array}\right.$  消去未知數(shù)y后，得到的方程組可能是（　�。�

• A． $\left\{\begin{array}{l}{7x+z=4}\\{5x-z=12}\end{array}\right.$
• B． $\left\{\begin{array}{l}{7x+z=4}\\{x-5z=8}\end{array}\right.$
• C． $\left\{\begin{array}{l}{7x-z=12}\\{x-5z=28}\\{\;}\end{array}\right.$
• D． $\left\{\begin{array}{l}{7x-z=4}\\{x-5z=12}\end{array}\right.$

Answer 1

分析 利用加減消元法解出方程組即可．

解答 解：$\left\{\begin{array}{l}{2x-3y+2z=5①}\\{x-2y+3z=-6②}\\{3x-y+z=3③}\end{array}\right.$，
③×3-①得，7x+z=4④，
③×2-②得，5x-z=12⑤，
由④⑤組成方程組得，$\left\{\begin{array}{l}{7x+z=4}\\{5x-z=12}\end{array}\right.$，
故選：A．

點(diǎn)評(píng) 本題考查的是二元一次方程組、三元一次方程組的解法，掌握加減消元法是解題的關(guān)鍵．