Question

11．解方程组：$\left\{\begin{array}{l}x-2y=3\\ 4{x^2}-12xy+9{y^2}=16\end{array}\right.$．

Answer 1

分析 由②得出（2x-3y）²=16，求出2x-3y=±4，把原方程组转化成两个二元一次方程组，求出方程组的解即可．

解答 解：$\left\{\begin{array}{l}{x-2y=3①}\\{4{x}^{2}-12xy+9{y}^{2}=16②}\end{array}\right.$
由②得：（2x-3y）²=16，
2x-3y=±4，
即原方程组化为$\left\{\begin{array}{l}{x-2y=3}\\{2x-3y=4}\end{array}\right.$和$\left\{\begin{array}{l}{x-2y=3}\\{2x-3y=-4}\end{array}\right.$，
解得：$\left\{\begin{array}{l}{{x}_{1}=-1}\\{{y}_{1}=-2}\end{array}\right.$，$\left\{\begin{array}{l}{{x}_{2}=-17}\\{{y}_{2}=-10}\end{array}\right.$，
即原方程组的解为：$\left\{\begin{array}{l}{{x}_{1}=-1}\\{{y}_{1}=-2}\end{array}\right.$，$\left\{\begin{array}{l}{{x}_{2}=-17}\\{{y}_{2}=-10}\end{array}\right.$．

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