Question

4．下列方程組：①$\left\{\begin{array}{l}{2x-y=1}\\{y=z+1}\end{array}\right.$②$\left\{\begin{array}{l}{x-2y=0}\\{y=3}\end{array}\right.$③$\left\{\begin{array}{l}{x-y=0}\\{2x+3y=5}\end{array}\right.$④$\left\{\begin{array}{l}{{x}^{2}+y=1}\\{x+2y=-1}\end{array}\right.$其中是二元一次方程組的有（　�。�

• A． 1個(gè)
• B． 2個(gè)
• C． 3個(gè)
• D． 4個(gè)

Answer 1

分析 利用二元一次方程組的定義判斷即可．

解答 解：①$\left\{\begin{array}{l}{2x-y=1}\\{y=z+1}\end{array}\right.$，不是；②$\left\{\begin{array}{l}{x-2y=0}\\{y=3}\end{array}\right.$，是；③$\left\{\begin{array}{l}{x-y=0}\\{2x+3y=5}\end{array}\right.$，是；④$\left\{\begin{array}{l}{{x}^{2}+y=1}\\{x+2y=-1}\end{array}\right.$，不是；
則其中是二元一次方程組的有2個(gè)．
故選B．

點(diǎn)評(píng) 此題考查了二元一次方程組的定義，熟練掌握二元一次方程組的定義是解本題的關(guān)鍵．