Question

16．解二元二次方程组$\left\{\begin{array}{l}{5{x}^{2}+2{y}^{2}+x-4y-6=0}\\{2{x}^{2}+{y}^{2}+x-2y-3=0}\end{array}\right.$．

Answer 1

分析 $\left\{\begin{array}{l}{5{x}^{2}+2{y}^{2}+x-4y-6=0}&{①}\\{2{x}^{2}+{y}^{2}+x-2y-3=0}&{②}\end{array}\right.$，①-②×2可得：x²-x=0，解得x，进而解得y．

解答 解：$\left\{\begin{array}{l}{5{x}^{2}+2{y}^{2}+x-4y-6=0}&{①}\\{2{x}^{2}+{y}^{2}+x-2y-3=0}&{②}\end{array}\right.$，
①-②×2可得：x²-x=0，解得x=0或1．
∴$\left\{\begin{array}{l}{x=0}\\{y=-1}\end{array}\right.$，$\left\{\begin{array}{l}{x=0}\\{y=3}\end{array}\right.$，$\left\{\begin{array}{l}{x=1}\\{y=0}\end{array}\right.$，$\left\{\begin{array}{l}{x=1}\\{y=2}\end{array}\right.$．
∴原方程组的解为：$\left\{\begin{array}{l}{x=0}\\{y=-1}\end{array}\right.$，$\left\{\begin{array}{l}{x=0}\\{y=3}\end{array}\right.$，$\left\{\begin{array}{l}{x=1}\\{y=0}\end{array}\right.$，$\left\{\begin{array}{l}{x=1}\\{y=2}\end{array}\right.$．

点评 本题考查了方程组的解法，考查了推理能力与计算能力，属于中档题．