Question

15．解二元二次方程组$\left\{\begin{array}{l}{{x}^{2}+2xy+3{y}^{2}-48x+4y-4=0}\\{2{x}^{2}+4xy+6{y}^{2}-99x+7y-6=0}\end{array}\right.$．

Answer 1

分析 $\left\{\begin{array}{l}{{x}^{2}+2xy+3{y}^{2}-48x+4y-4=0}&{①}\\{2{x}^{2}+4xy+6{y}^{2}-99x+7y-6=0}&{②}\end{array}\right.$，②-①×2可得：y=2-3x，代入①化为：11x²-46x+8=0，解得x，进而解得y．

解答 解：$\left\{\begin{array}{l}{{x}^{2}+2xy+3{y}^{2}-48x+4y-4=0}&{①}\\{2{x}^{2}+4xy+6{y}^{2}-99x+7y-6=0}&{②}\end{array}\right.$，
②-①×2可得：y=2-3x，代入①化为：11x²-46x+8=0，解得x=$\frac{2}{11}$，x=4．
∴$\left\{\begin{array}{l}{x=\frac{2}{11}}\\{y=\frac{16}{11}}\end{array}\right.$，或$\left\{\begin{array}{l}{x=4}\\{y=-10}\end{array}\right.$．
∴原方程组的解为：$\left\{\begin{array}{l}{x=\frac{2}{11}}\\{y=\frac{16}{11}}\end{array}\right.$，或$\left\{\begin{array}{l}{x=4}\\{y=-10}\end{array}\right.$．

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．