Question

若方程组x+y=3-2m/5 2x+3y=4-3m的解x、y满足x<y,求m的取值范围

Answer 1

x+y=3-2m/5 (1)
2x+3y=4-3m (2)
(2)-2*(1)得:
y=4-3m-6+4m/5=-11m/5-2
x=3-2m/5+11m/5+2=9m/5+5
x<y
即
9m/5+5<-11m/肢旁5-2
4m<-7
m<-7/4
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Answer 2

x+y=3-2m/5 (1)
2x+3y=4-3m (2)

(1)×5-(2)×2
5x+5y-4x-6y=15-2m-8+6m
x-y=4m+7
x<y则氏兆x-y<迟虚0
4m+7<码核燃0
m<-7/4

Answer 3

解方程组x+y=3-2m/稿汪5 2x+3y=4-3m得：粗昌
x=5+9m/5,y=-2-11m/岩敬扒5
5+9m/5<-2-11m/5
m<-7/4

Answer 4

先解方程，求出x、y的值，即：x=5+9m/5 y=-2-11m/5
因数拦携x<y，所以：5+9m/5<衡陪-2-11m/5，求得：薯伏m<-7/4