The sum of three numbers is 10. two times the second number minus the first number is equal to 12. the first number minus the second number plus twice the third number equals 7. find the numbers.

let a=1st number
let b=2nd number
let c=3rd number
then
a+b+c=10
2b-a=12
a-b+2c=7

a+b+c=10
-a+b=12
add
3b+c=22

-a+2b=12
a-b+2c=7
add

b+2c=19

3b+c=22
b+2c=19

3b+c=22
3b+6c=57
subtract
-5c=-35
c=7
b=-2c+19=5
a=2b-12=-2

a=1 number=-2
b=2nd number=5
c=3rd number=7 

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