已知实数x、y满足
2x-3y=4
x=2+1.5y
x的最大值
x＜2+1.5×2
x＜5
k=x+y的最大值
x+y＜5+2=7

并且x≥-1，y＜2，现有k=x＋y，则k的取值范围是_____求过程

已知实数x、y满足
2x-3y=4
x=2+1.5y
x的最大值
x＜2+1.5×2
x＜5
k=x+y的最大值
x+y＜5+2=7

2x-3y=4
x=2+1.5y
x的最小值是-1
y的最小值
y=(-1-2)/1.5=-2

k=x+y的最小值
x+y=-1-2=-3
-3＜k≦7

先把2x-3y=4变形得到y=3分之1（2x-4），由y＜2得到3分之1（2x-4）＜2，解得x＜5，所以x的取值范围为-1≤x＜5，再用x变形k得到k=3分之1x+3分之4，然后利用一次函数的性质确定k的范围．
解：∵2x-3y=4，
∴y=3分之1（2x-4），
∵y＜2，
∴3分之1（2x-4）＜2，解得x＜5，
又∵x≥-1，
∴-1≤x＜5，
∵k=x-3分之1（2x-4）=3分之1x+3分之4，
当x=-1时，k=×3分之1（-1）+3分之4=1；
当x=5时，k=3分之1×5+3分之4=3，
∴1≤k＜3．
故答案为：1≤k＜3．

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