题目内容

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

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.

练习册系列答案
相关题目

违法和不良信息举报电话:027-86699610 举报邮箱:58377363@163.com

精英家教网