Question

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

Answer 1

分析 首先把原来的方程组化成二元一次方程组，然后应用加减法，求出方程组的解是多少即可．

解答 解：∵$\left\{\begin{array}{l}{y-2x=6}\\{4{x}^{2}+4xy+{y}^{2}=4}\end{array}\right.$，
∴$\left\{\begin{array}{l}{y-2x=6}\\{2x+y=2}\end{array}\right.$或$\left\{\begin{array}{l}{y-2x=6}\\{2x+y=-2}\end{array}\right.$
解得$\left\{\begin{array}{l}{x=-1}\\{y=4}\end{array}\right.$或$\left\{\begin{array}{l}{x=-2}\\{y=2}\end{array}\right.$．

点评 此题主要考查了高次方程的求解方法，要熟练掌握，注意解高次方程一般要降次，即把它转化成二次方程或一次方程．