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.
Listed in order from smallest to largest, the numbers are
and
DONE
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Someone else had this problem, so I copied the

the first number is -2
the 2nd is 5
the third is 7

Equations:
X+y+z=10
2y-x=12
x-y+2z=7

Try to eliminate y and z by substitution

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)

substitute 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

substitute 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

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