If the equations MX + 2ny = 4, x + y = 1 and X-Y = 3, NX + (m-1) y = 3 have the same solution

If the equations MX + 2ny = 4, x + y = 1 and X-Y = 3, NX + (m-1) y = 3 have the same solution

If x + y = 1, X-Y = 3, x = 2, y = - 1
Substituting MX + 2ny = 4, NX + (m-1) y = 3
The result is: 2m-2n = 4 (m-n = 2)
2n-m+1=3(2n-m=2)
Simulink: n = 4, M = 6

We know the equations {MX + 2ny = 4 x + y = 1 and {x-6y = 3 NX + (m-1) y = 3 for X y
Finding the value of M and N with the same solution

x+y=1
x-6y=3
subtract
7y=-2
y=-2/7
x=9/7
Substituting
9m/7-4n/7=4
That is, 9m-4n = 28 (1)
9n/7-3(m-1)/7=3
That is 3N-M = 6 (2)
So m = 3n-6
Then 27n-54-4n = 28
23n=82
therefore
n=82/23
m=108/23

It is known that the system of equations MX + 2ny = 4, x + y = 1 and X-Y = 3, NX + (m-1) y = 3 have the same solution. Find the value of M, n

Solution x + y = 1 (1)
x-y=3(2）
We get: x = 2, y = - 1
Taking x = 2, y = - 1 into MX + 2ny = 4 and NX + (m-1) y = 3, we get that
2m-2n=4
2n-(m-1)=3
We get: M = 6, n = 4