Question

1．已知关于x，y的方程组$\left\{\begin{array}{l}{3x-y=k}\\{2x+3y=2k-1}\end{array}\right.$的解满足不等式$\left\{\begin{array}{l}{x-4y≥0}\\{5x+2y≥-2}\end{array}\right.$，求满足条件的k的整数值．

Answer 1

分析 将k看做已知数表示出x与y，代入已知不等式组即可求出k的范围，即可确定整数值．

解答 解：方程组$\left\{\begin{array}{l}{3x-y=k}\\{2x+3y=2k-1}\end{array}\right.$
解得：$\left\{\begin{array}{l}{x=\frac{5k-1}{11}}\\{y=\frac{4k-3}{11}}\end{array}\right.$，
把$\left\{\begin{array}{l}{x=\frac{5k-1}{11}}\\{y=\frac{4k-3}{11}}\end{array}\right.$代入不等式$\left\{\begin{array}{l}{x-4y≥0}\\{5x+2y≥-2}\end{array}\right.$得：
$\left\{\begin{array}{l}{-11k+12≤0}\\{3k+11≥0}\end{array}\right.$，
解得：$\left\{\begin{array}{l}{k≤\frac{12}{11}}\\{k≥-\frac{11}{3}}\end{array}\right.$，
∴$-\frac{11}{3}≤k≤\frac{12}{11}$．
∴k=-3，-2，-1，0，1．

点评 此题考查了二元一次方程组的解，方程组的解即为能使方程组中两方程成立的未知数的值，解决本题的关键是求出方程组的解，列出不等式组．