simultaneous equations sub method?

m-2n=3

m=3+2n


5m+2n=3

5(3+2n)+2n=3

15+10n+2n=3

15+12n=3

12n=3-15

12n=-12

n=-12/12

n=-1

m - 2n = 3
5m + 2n = 3

m = 3 + 2n
5m = 5(3 + 2n)

5m + 2n = 3
5(3 + 2n) + 2n = 3
15 + 10n + 2n = 3
10n + 2n = 3 - 15
12n = -12
n = -12/12
n = -1

m - 2n = 3
m - 2(-1) = 3
m - (-2) = 3
m + 2 = 3
m = 3 - 2
m = 1

∴ m = 1 , n = -1

i got 5/6 because you cannel the opposite signs (-2n and +2n) it left me with m=3 and 6m=5. you then have to add them (6m-5) it gave me 5/6

i think u meant 2n not sn

m-2n=3 equation 1
5m+2n=3 equation 2

rearrange eq 1
m=3+2n equation 3
sub eq 3 into eq 2

5(3+2n)+2n=3
15+10n+2n=3
15+12n=3
12n=-12
n=-1

sub n=-1 into eq 1
m-2n=3
m+2=3
m=1

check with eq 2
5m+2n=3
5-2=3
3=3

therefore solution is n=-1 m=1

taking a wild guess that you intended to type

m-2n=3
5m+2n=3

Personally I'd add to eliminate n and get m=1 then sub that in to get n=-1

Straight substitution method would have you 3+2n in place of m in the second equation giving 5(3+2n)+2n=3 etc...

First equation will not be m - sn = 3 will it ??????