Question

12．方程組$\left\{\begin{array}{l}{x=2y+1}\\{2x-y=11}\end{array}\right.$的解是$\left\{\begin{array}{l}{x=7}\\{y=3}\end{array}\right.$．

Answer 1

分析 方程組利用代入消元法求出解即可．

解答 解：$\left\{\begin{array}{l}{x=2y+1①}\\{2x-y=11②}\end{array}\right.$，
把①代入②得：2（2y+1）-y=11，
去括號(hào)得：4y+2-y=11，
移項(xiàng)合并得：3y=9，
解得：y=3，
把y=3代入①得：x=7，
則方程組的解為$\left\{\begin{array}{l}{x=7}\\{y=3}\end{array}\right.$
故答案為：$\left\{\begin{array}{l}{x=7}\\{y=3}\end{array}\right.$

點(diǎn)評(píng) 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．