题目内容
17.整数k取何值时,方程组$\left\{\begin{array}{l}{3x+y=2k}\\{x-2y=-3}\end{array}\right.$的解满足条件:x<1且y>1?分析 先求出方程组的解,根据题意得出不等式组,求出不等式组的解集,即可得出答案.
解答 解:解方程组$\left\{\begin{array}{l}{3x+y=2k}\\{x-2y=-3}\end{array}\right.$得:$\left\{\begin{array}{l}{x=\frac{4k-3}{7}}\\{y=\frac{2k+9}{7}}\end{array}\right.$,
∵x<1且y>1,
∴$\left\{\begin{array}{l}{\frac{4k-3}{7}<1}\\{\frac{2k+9}{7}>1}\end{array}\right.$,
解得:-1<k<$\frac{5}{2}$,
整数k为0,1,2,
即当整数k取0或1或2时,方程组$\left\{\begin{array}{l}{3x+y=2k}\\{x-2y=-3}\end{array}\right.$的解满足条件:x<1且y>1.
点评 本题考查了二元一次方程组的解得应用,能根据题意求出关于k的不等式组是解此题的关键.
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