x+y+z=10
2y-x=12
x-y+2z=7
k
so we do math suff
hmm
ok, so try to eliminate y and z by subsitution
solve for everything in terms of x in last 2 equations
2y-x=12
2y=12+x
y=6+(1/2)x
x-y+2z=7
2z=7+y-x
z=(1/2)(7+y-x)
and y=6+(1/2)x
z=(1/2)(7+6+(1/2)x-x)
z=(1/2)(13+(-1/2)x)
sub those
x+y+z=10
x+6+(1/2)x+(1/2)(13+(-1/2)x)=10
x+6+(1/2)x+(13/2)+(-1/4)x=10
(4/4)x+(12/2)+(2/4)x+(13/2)+(-1/4)x=10
(5/4)x+(25/2)=10
minus (25/2) from both sides
(5/4)x=(20/2)-(25/2)
(5/4)x=-5/2
times both sides by 4
5x=-10
divide both sides by 5
x=-2
sub back
y=6+(1/2)x
y=6+(1/2)(-2)
y=6-1
y=5
z=(1/2)(13+(-1/2)x))
z=(1/2)(13+(-1/2)(-2))
z=(1/2)(13+1)
z=(1/2)(14)
z=7
the first number is -2
the 2nd is 5
the third is 7