Question

5．二元一次方程組$\left\{\begin{array}{l}{x+y=2}\\{x-y=0}\end{array}\right.$的解是（　�。�

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

Answer 1

分析 方程組利用加減消元法求出解即可．

解答 解：$\left\{\begin{array}{l}{x+y=2①}\\{x-y=0②}\end{array}\right.$，
①+②得：2x=2，即x=1，
①-②得：2y=2，即y=1，
則方程組的解為$\left\{\begin{array}{l}{x=1}\\{y=1}\end{array}\right.$，
故選B

點(diǎn)評 此題考查了解二元一次方程組，熟練掌握運(yùn)算法則是解本題的關(guān)鍵．