The sum of three numbers is -2.

1) a + b + c = -2

The sum of three times the first number, twice times the second number, and the third number is 9.

2) 3a + 2b + c = 9

Subtract 1) from 2)

3) 2a + b = 11

The difference between the second number and half the third number is 10. Find the numbers.

4) Two possible cases:

Case 1: b - c/2 = 10 if b > c/2

Case 2: c/2 - b = 10 if b < c/2

In either case the difference between the two terms is 10

Case 1) b > c/2

Multiply 4) by 2 and solve for c

2b - c = 20

c = 2b-20

Substitute in 1)

a + b + 2b-20 = -2

5) a + 3b = 18

Use 3) and 5) to find a and b

2a + b = 11 → 6a + 3b = 33

a + 3b = 18

6a -a + 3b - 3b = 33-18

5a = 15

a = 3

Use 3) to find b

2(3) + b = 11

b = 5

c = 2b-20 = -10

Ans: (a, b, c) = (3, 5, -10)

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Case 2) b < c/2

Multiply 4) by 2 and solve for c

c - 2b = 20

c = 2b+20

Substitute in 1)

a + b + 2b+20 = -2

5) a + 3b = -22

Use 3) and 5) to find a and b

2a + b = 11 → 6a + 3b = 33

a + 3b = -22

6a -a + 3b - 3b = 33+18

5a = 55

a = 11

Use 3) to find b

2(11) + b = 11

b = -11

c = 2b+20 = -2

Ans: (a, b, c) = (11, -11, -2)

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There are two sets of numbers which satisfy the problem statement:

Solution 1) 3, 5, -10

Solution 2) 11, -11, -2

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Recall that the "difference" between two numbers a and b is always a positive number:

|a - b| = |b - a| ≥ 0

This leads to two solutions which satisfy the problem statement.