Knowing that the solution of the equation 2x my 4, x 4y 8 is a positive integer, find the value of m

2x+my-2x-8y=4-16

8-m)y=12

y=(8-m)/12>0

x+(8-m)/3=8

x=(16+m)/3>0

So. 8-m must be a positive integer multiple of 12 (1).

16+m must be a positive integer multiple of 3 (2).

Start with condition (2).

m=-1 is not true.

m=-4 holds.

m=-7 is not true.

m=-10 is not true.

m=-13, -16 are not true.

So only m=-4 also meets conditions (1) and (2).

Then m=-4

From the second equation, we can see that x can only be 1, 2, 4

x=1 can be set to be checked by the second formula, which is not conformed.

When x=2, it is tested by the second formula and does not meet.

When x=4, y=1

Substituting the first equation yields, m=-4

x = 8-4y and 0 is equal to -4 when m = 0, y (not a positive integer).

When m=-1, y (not a positive integer).

m=Not true.

m=-4,y=1,x=4

When m= is not true.

When m=2, y=2, x=0 (not a positive integer).

When m=4, y=3, x=-4 (not a positive integer).

When m=5, y=4, x=-8 (not a positive integer).

When m=6, y=6, x=-16 (not a positive integer) When m=7, y=12, x=-40 (not a positive integer) the final result: m=-4, y=1, x=4

x=8-4y

x y is a positive integer, then y can only wait for 1

Since y>=2, then x<=0

y=1, then x=4

8+m=4m=-4

2x+my-2x-8y=4-16

8-m)y=12

y=(8-m)/12>0

x+(8-m)/3=8

x=(16+m)/3>0

So. 8-m must be a positive integer multiple of 12 (1) and 16+m must be a positive integer multiple of 3 (2).

Start with condition (2).

m=-1 does not hold the beat.

m=-4 holds.

m=-7 is not true.

m=-10 is not true.

m=-13, -16 are not Hongnali.

So only m=-4 also meets conditions (1) and (2).

Then m=-4

2x+my=4………1）

x+4y=8………Fight the Chamber....（2）

From (2), get:

x=8-4y

If the solution is a positive integer, then 8-4y>0

y<2 so it can only be y=1

x=8-4=4

Substituting equation (1), we get:

2×4+m=4

m=-4 In summary, the value of m is -4, and the solution of the square and Bicheng groups is x=4, y=1

4x+2my=16 (1)

4x+4y=8 (2)

1)-(2), got.

2my-4y=8

m-2)y=4

y=4/(m-2)

Since y is an integer, then m-2 = 1, 2, 4

m=3,4,6 ,1,0,-2

When the ruler m=3, y=4, x=-2

When m=4, y=2, x=0

When m=6, y=1, x=1

When m=1, y=-4, x=6

are full of buried and full of inscriptions.

Therefore, the positive integer m=1,4 or the remaining 6 in the bending position

2x+my=4………1）

x+4y=8………Fight the Chamber....（2）

From (2), get:

x=8-4y

If the solution is a positive integer, then 8-4y>0

y<2 so it can only be y=1

x=8-4=4

Substituting equation (1), we get:

2×4+m=4

m=-4 In summary, the value of m is -4, and the solution of the square and Bicheng groups is x=4, y=1

2x+my=4

x+4y=8

Simultaneous formulas give x=(16-8m) (8-m), y=12 (8-m)1) are both positive.

then (16-8m) (8-m)>0 and 12 (8-m)>0 can be known m

2x+my=4

x+4y=8

Simultaneous formulas give x=(16-8m) (8-m), y=12 (8-m)1) are both positive.

Then (16-8m) (8-m)>0 and 12 (8-m)>0 can be seen that m<2 stove feasts with m<2 are all positive luck.

2) All are positive integers.

Since y=12 (8-m) is a positive integer, the value of y can only be from m<2 when both are positive numbers to know that y=1 can be satisfied (when y=2 m=2, x=(16-16) 6=0, not x>0).

When y=1, m=-4 x=(16+32) (12)=4 satisfies x to be a positive integer.

So when m=-4, the full hidden silver foot is a positive integer.

System of equations 2x+my=4 (1); x+4y=8（2）；Change the loss of equation (2) to x=8-4y(3).

Substituting Eq. (1) yields 16-8y+my=4 to Y=12 (8-m) (4), and substituting Eq. (4) into Eq. (3) yields x=8-12 (8-m)=(52-8m) (8-m)(5).

If the condition (1) is satisfied, then the denominator of equation (4) can be obtained by m<8, and the numerator of equation (5) is obtained by m<, and the condition (1) is satisfied when m"; To satisfy condition (2), then on the basis of condition (1), equation (4) must be 8 y 0;y can only be an integer from 1 to 7, then m can only be , -4.

2x+my=4………1) Infiltration.

x+4y=8………2）

From (2), get:

x=8-4y

If the solution is a positive integer, then 8-4y>0

y "Song Search 2, so I can only shout that the calendar is y=1."

x=8-4=4

Substituting equation (1), we get:

2×4+m=4

m=-4In summary, the value of m is -4, and the solution of the system of equations is x=4, y=1

x+4y=8 (1)

2x+my=4 (2)

8-m)y=12

y=12/(8-m) (3)

Substituting (1) gives x=8-12 (8-m)=(52-8m) (8-m) (4).

Because the solution of the system of equations is a positive integer.

So from (3) 8-m is a factor of 12.

When 8-m=1, y=12 m=7 is substituted for (4), x=-4 is not in line with the topic, so discard.

When 8-m=2, y=6 m=6 is substituted for (4), x=2 is eligible.

When 8-m=3, y=4 m=5 is substituted for (4), x=4 is eligible.

When 8-m=4, y=3 m=4 is substituted into (4), x=5 is eligible.

When 8-m=6, y=2 m=2 is substituted for (4), x=6 is eligible.

When 8-m=12, y=1 m=-4 is substituted for (4), x=7 is eligible.

So m=6 or 5 or 4 or 2 or -4

The solution of this system of equations is x=4,y=1

Substituting x and y into the second equation gives 2 4 + m 1 = 4, and the solution gives m = -4