Question

11．解方程组：$\left\{\begin{array}{l}{x+2y=6}\\{{x}^{2}-3xy+2{y}^{2}=0}\end{array}\right.$．

Answer 1

分析 由②得出x-2y=0，x-y=0，这样转化成两个方程组，求出方程组的解即可．

解答 解：$\left\{\begin{array}{l}{x+2y=6①}\\{{x}^{2}-3xy+2{y}^{2}=0②}\end{array}\right.$
由②得，x-2y=0，x-y=0，
原方程组化为$\left\{\begin{array}{l}{x+2y=6}\\{x-y=0}\end{array}\right.$或$\left\{\begin{array}{l}{x+2y=6}\\{x-2y=0}\end{array}\right.$，
解得：$\left\{\begin{array}{l}{{x}_{1}=2}\\{{y}_{1}=2}\end{array}\right.$，$\left\{\begin{array}{l}{{x}_{2}=3}\\{{y}_{2}=\frac{3}{2}}\end{array}\right.$，
∴原方程组的解是 $\left\{\begin{array}{l}{{x}_{1}=2}\\{{y}_{1}=2}\end{array}\right.$，$\left\{\begin{array}{l}{{x}_{2}=3}\\{{y}_{2}=\frac{3}{2}}\end{array}\right.$．

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