Three math problems about equation in grade six &To solve the equations {- x + 2Y + 3Z = 12 & nbsp; {2x-2y + 5Z = 13 & nbsp; {3x + 2Y + Z = 10 & nbsp; Solve the system of equations {2 (x + y) - 3 (X-Y) = 1 & nbsp; {6 (x + y) + 7 (X-Y) = 51 & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; remove the brackets and make it easier Given that the system of equations {2x-y = 0 & nbsp; {2x + y = m and the system of equations {X-Y = 5 & nbsp; {2x-y = n-1 have a set of common solutions, the values of M and N are obtained   ,,

Three math problems about equation in grade six &To solve the equations {- x + 2Y + 3Z = 12 & nbsp; {2x-2y + 5Z = 13 & nbsp; {3x + 2Y + Z = 10 & nbsp; Solve the system of equations {2 (x + y) - 3 (X-Y) = 1 & nbsp; {6 (x + y) + 7 (X-Y) = 51 & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; remove the brackets and make it easier Given that the system of equations {2x-y = 0 & nbsp; {2x + y = m and the system of equations {X-Y = 5 & nbsp; {2x-y = n-1 have a set of common solutions, the values of M and N are obtained   ,,

1.-x+2y+3z=12 ①
2x-2y+5z=13 ②
3x+2y+z=10 ③
① + 2 gives x + 8Z = 25 4
② + 3 is 5x + 6Z = 23 5
④ 5, then 5x + 40z = 125 6
⑥ (5) 34z = 102
z=3 ⑦
⑦ Substituting 4 gives x + 8 × 3 = 25
x+24=25
x=25-24
x=1 ⑧
⑦ 3 + 2Y + 3 = 10 is obtained by substituting 8 with 3
2y+6=10
2y=10-6
2y=4
y=2
The solution of the equations is x = 1, y = 2, z = 3
2.2(x+y)-3(x-y)=1 ①
6(x+y)+7(x-y)=51 ②
① It is reduced to 2x + 2y-3x + 3Y = 1
x-5y=-1
x=5y-1 ③
② It is reduced to 6x + 6y + 7x-7y = 51
13x-y = 51 (4)
③ Substituting 4 gives 13 (5y-1) - y = 51
65y-13-y=51
65y-y=51+13
64y=64
y=1 ⑤
⑤ Substituting 3 gives x = 5-1
x=4
The solution of the equations is x = 4, y = 1
3. Calculate the value of X and Y first
2x-y=0 ①
x-y=5 ②
① (2) get x = - 5 (3)
③ Substituting into 2, we get - 5-y = 5
-y=5+5
y=-10④
Substituting (3) and (4)
2x+y=m
2×(-5)-10=m
m=-10-10
m=-20
Substituting (3) and (4)
2x-y=n-1
2×(-5)+10=n-1
-10+10=n-1
n=1
The solution is m = - 20, n = 1