题目内容
16．方程组$\left\{\begin{array}{l}{x+y=3}\\{x-2y=-3+a}\end{array}\right.$的解满足$\left\{\begin{array}{l}{x≥0}\\{y≥0}\end{array}\right.$，求a的取值范围．分析 解方程组可得关于a的不等式组，解不等式组可得．
解答 解：解方程组$\left\{\begin{array}{l}{x+y=3}\\{x-2y=-3+a}\end{array}\right.$可得$\left\{\begin{array}{l}{x=1+\frac{a}{3}}\\{y=2-\frac{a}{3}}\end{array}\right.$，
∵$\left\{\begin{array}{l}{x≥0}\\{y≥0}\end{array}\right.$，∴$\left\{\begin{array}{l}{1+\frac{a}{3}≥0}\\{2-\frac{a}{3}≥0}\end{array}\right.$，解得-3≤a≤6．
点评 本题考查不等式的解法，涉及方程组的解集，属基础题．
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