Question

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

Answer 1

分析 由于2x+y的符號不確定，故應分2x+y≥0與2x+y＜0兩種情況進行討論．

解答 解：原方程組可化為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$\left\{\begin{array}{l}{x=-4}\\{y=-6}\end{array}\right.$．
故原方程組的解為$\left\{\begin{array}{l}{x=4}\\{y=-2}\end{array}\right.$或$\left\{\begin{array}{l}{x=-4}\\{y=-6}\end{array}\right.$．

點評 本題考查的是解二元一次方程組，熟知解二元一次方程組的加減消元法和代入消元法是解答此題的關(guān)鍵．