Question

2．二元一次方程組$\left\{\begin{array}{l}{x+2y=2}\\{2x+y=-2}\end{array}\right.$的解是（　�。�

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

Answer 1

分析 ①×2-②得出3y=6，求出y=2，把y的值代入①求出x即可．

解答 解：$\left\{\begin{array}{l}{x+2y=2①}\\{2x+y=-2②}\end{array}\right.$
①×2-②得：3y=6，
解得：y=2，
把y=2代入①得：x+4=2，
解得：x=-2，
即方程組$\left\{\begin{array}{l}{x+2y=2}\\{2x=y=-2}\end{array}\right.$的解為$\left\{\begin{array}{l}{x=-2}\\{y=2}\end{array}\right.$，
故選B．

點(diǎn)評(píng) 本題考查了解二元一次方程組的應(yīng)用，能把二元一次方程組轉(zhuǎn)化成一元一次方程是解此題的關(guān)鍵．