Question

19．解方程组：$\left\{\begin{array}{l}{x+|2x+y|=10，①}\\{x-2y=8，②}\end{array}\right.$．

Answer 1

分析 由于2x+y的符号不确定，故应分2x+y≥0与2x+y＜0两种情况进行讨论．

解答 解：原方程组可化为①$\left\{\begin{array}{l}{x+2x+y=10}\\{x-2y=8}\end{array}\right.$或②$\left\{\begin{array}{l}{x-2x-y=10}\\{x-2y=8}\end{array}\right.$，
解①得$\left\{\begin{array}{l}{x=4}\\{y=-2}\end{array}\right.$，解②得$\left\{\begin{array}{l}{x=-4}\\{y=-6}\end{array}\right.$．
故原方程组的解为$\left\{\begin{array}{l}{x=4}\\{y=-2}\end{array}\right.$或$\left\{\begin{array}{l}{x=-4}\\{y=-6}\end{array}\right.$．

点评 本题考查的是解二元一次方程组，熟知解二元一次方程组的加减消元法和代入消元法是解答此题的关键．