Question

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

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化為：11x²-46x+8=0，解得x，進(jìn)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化為：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$\left\{\begin{array}{l}{x=4}\\{y=-10}\end{array}\right.$．
∴原方程組的解為：$\left\{\begin{array}{l}{x=\frac{2}{11}}\\{y=\frac{16}{11}}\end{array}\right.$，或$\left\{\begin{array}{l}{x=4}\\{y=-10}\end{array}\right.$．

點(diǎn)評(píng) 本題考查了方程組的解法，考查了推理能力與計(jì)算能力，屬于中檔題．