Question

2．已知$\left\{{\begin{array}{l}{{x_1}=3}\\{{y_1}=-2}\end{array}}\right.$是方程组$\left\{{\begin{array}{l}{{x^2}+{y^2}=m}\\{x+y=n}\end{array}}\right.$的一组解，那么此方程组的另一组解是$\left\{\begin{array}{l}{{x}_{2}=-2}\\{{y}_{2}=3}\end{array}\right.$．

Answer 1

分析 先将$\left\{{\begin{array}{l}{{x_1}=3}\\{{y_1}=-2}\end{array}}\right.$代入方程组$\left\{{\begin{array}{l}{{x^2}+{y^2}=m}\\{x+y=n}\end{array}}\right.$中求出m、n的值，然后再求方程组的另一组解．

解答 解：将$\left\{{\begin{array}{l}{{x_1}=3}\\{{y_1}=-2}\end{array}}\right.$代入方程组$\left\{{\begin{array}{l}{{x^2}+{y^2}=m}\\{x+y=n}\end{array}}\right.$中得：$\left\{\begin{array}{l}{9+4=m}\\{3-2=n}\end{array}\right.$，
解得：m=13，n=1，
则方程组变形为：$\left\{\begin{array}{l}{{x}^{2}+{y}^{2}=13}\\{x+y=1}\end{array}\right.$，
由x+y=1得：x=1-y，
将x=1-y代入方程x²+y²=13中可得：y²-y-6=0，
解得y=3或y=-2，
将y=3代入x+y=1中可得：x=-2，
所以方程的另一组解为：$\left\{\begin{array}{l}{{x}_{2}=-2}\\{{y}_{2}=3}\end{array}\right.$．
故答案为：$\left\{\begin{array}{l}{{x}_{2}=-2}\\{{y}_{2}=3}\end{array}\right.$．

点评 本题考查了高次方程，二元一次方程组的解法，熟记解二元一次方程的解法是解题的关键．