Question

3．解方程組：
$\left\{\begin{array}{l}{|x+y|=1}\\{|x|+2|y|=3}\end{array}\right.$．

Answer 1

分析 先去絕對值符號，再求出方程組的解即可．

解答 解：當x≥0，y≥0時，原方程組可化為$\left\{\begin{array}{l}{x+y=1}\\{x+2y=3}\end{array}\right.$①或$\left\{\begin{array}{l}{x+y=-1}\\{x+2y=3}\end{array}\right.$②；
當x＞0，y＜0時，原方程組可化為$\left\{\begin{array}{l}{x+y=1}\\{x-2y=3}\end{array}\right.$③或$\left\{\begin{array}{l}{x+y=-1}\\{x-2y=3}\end{array}\right.$④；
當x＜0，y≥0時，原方程組可化為$\left\{\begin{array}{l}{x+y=1}\\{-x+2y=3}\end{array}\right.$⑤或$\left\{\begin{array}{l}{x+y=-1}\\{-x+2y=3}\end{array}\right.$⑥；
當x＜0，y＜0時，原方程組可化為$\left\{\begin{array}{l}{x+y=1}\\{-x-2y=3}\end{array}\right.$⑦或$\left\{\begin{array}{l}{x+y=-1}\\{-x-2y=3}\end{array}\right.$⑧．
解①得$\left\{\begin{array}{l}{x=-1}\\{y=2}\end{array}\right.$（不合題意）；解②$\left\{\begin{array}{l}{x=-5}\\{y=4}\end{array}\right.$（不合題意）；解③得$\left\{\begin{array}{l}{x=\frac{5}{3}}\\{y=-\frac{2}{3}}\end{array}\right.$；解④得$\left\{\begin{array}{l}{x=\frac{1}{3}}\\{y=-\frac{4}{3}}\end{array}\right.$；
解⑤得$\left\{\begin{array}{l}{x=-\frac{1}{3}}\\{y=\frac{4}{3}}\end{array}\right.$；解⑥得$\left\{\begin{array}{l}{x=-\frac{5}{3}}\\{y=\frac{2}{3}}\end{array}\right.$；解⑦得$\left\{\begin{array}{l}{x=5}\\{y=-4}\end{array}\right.$（不合題意）；解⑧得$\left\{\begin{array}{l}{x=1}\\{y=-2}\end{array}\right.$（不合題意）．
故原方程組的解為：$\left\{\begin{array}{l}{x=\frac{5}{3}}\\{y=-\frac{2}{3}}\end{array}\right.$或$\left\{\begin{array}{l}{x=\frac{1}{3}}\\{y=-\frac{4}{3}}\end{array}\right.$或$\left\{\begin{array}{l}{x=-\frac{1}{3}}\\{y=\frac{4}{3}}\end{array}\right.$或$\left\{\begin{array}{l}{x=-\frac{5}{3}}\\{y=\frac{2}{3}}\end{array}\right.$．

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鍵．