Question

6．解方程组：$\left\{\begin{array}{l}{{x}^{2}-4xy+{3y}^{2}=0①}\\{2x+y=21②}\end{array}\right.$．

Answer 1

分析 把①变形得出x-3y=0，x-y=0，则原方程组化为两个二元一次方程组，求出方程组的解即可．

解答 解：由①得：（x-3y）（x-y）=0，
x-3y=0，x-y=0，
则原方程组化为：$\left\{\begin{array}{l}{x-3y=0}\\{2x+y=21}\end{array}\right.$或$\left\{\begin{array}{l}{x-y=0}\\{2x+y=21}\end{array}\right.$
解得：$\left\{\begin{array}{l}{{x}_{1}=9}\\{{y}_{1}=3}\end{array}\right.$，$\left\{\begin{array}{l}{{x}_{2}=7}\\{{y}_{2}=7}\end{array}\right.$
∴原方程组的解为：$\left\{\begin{array}{l}{{x}_{1}=9}\\{{y}_{1}=3}\end{array}\right.$，$\left\{\begin{array}{l}{{x}_{2}=7}\\{{y}_{2}=7}\end{array}\right.$．

点评 本题考查了高次方程组，能把高次方程组转化成二元一次方程组是解此题的关键．