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已知方程组x+my=11x+3=2y有正整数解,则整数m的值为.

题目详情
x+my=11
x+3=2y
有正整数解,则整数m的值为______.
x+my=11
x+3=2y
x+my=11
x+3=2y
x+my=11
x+3=2y
x+my=11
x+3=2y
x+my=11x+my=11x+3=2yx+3=2y
▼优质解答
答案和解析
方程组
x+my=11
x+3=2y

∴x+my-x-3=11-2y,
解得:(m+2)y=14,
y=
14
m+2

∵方程组有正整数解,
∴m+2>0,m>-2,
又x=
22-3m
m+2

故22-3m>0,
解得:m<
22
3

故-2<m<
22
3
,整数m只能取-1,0,1,2,3,4,5,6,7.
又x,y均为正整数,
∴只有m=-1或0或5符合题意.
故答案为:-1或0或5.
x+my=11
x+3=2y
x+my=11
x+3=2y
x+my=11
x+3=2y
x+my=11
x+3=2y
x+my=11x+my=11x+my=11x+3=2yx+3=2yx+3=2y,
∴x+my-x-3=11-2y,
解得:(m+2)y=14,
y=
14
m+2

∵方程组有正整数解,
∴m+2>0,m>-2,
又x=
22-3m
m+2

故22-3m>0,
解得:m<
22
3

故-2<m<
22
3
,整数m只能取-1,0,1,2,3,4,5,6,7.
又x,y均为正整数,
∴只有m=-1或0或5符合题意.
故答案为:-1或0或5.
14
m+2
141414m+2m+2m+2,
∵方程组有正整数解,
∴m+2>0,m>-2,
又x=
22-3m
m+2

故22-3m>0,
解得:m<
22
3

故-2<m<
22
3
,整数m只能取-1,0,1,2,3,4,5,6,7.
又x,y均为正整数,
∴只有m=-1或0或5符合题意.
故答案为:-1或0或5.
22-3m
m+2
22-3m22-3m22-3mm+2m+2m+2,
故22-3m>0,
解得:m<
22
3

故-2<m<
22
3
,整数m只能取-1,0,1,2,3,4,5,6,7.
又x,y均为正整数,
∴只有m=-1或0或5符合题意.
故答案为:-1或0或5.
22
3
222222333,
故-2<m<
22
3
,整数m只能取-1,0,1,2,3,4,5,6,7.
又x,y均为正整数,
∴只有m=-1或0或5符合题意.
故答案为:-1或0或5.
22
3
222222333,整数m只能取-1,0,1,2,3,4,5,6,7.
又x,y均为正整数,
∴只有m=-1或0或5符合题意.
故答案为:-1或0或5.