Final answer:
Using system of equations, the three integers that satisfy the conditions that their sum is 18, the third is twice the second, and the first is six more than the second, are found to be 9, 3, and 6 respectively.
Step-by-step explanation:
To solve the problem of finding the three integers a, b, and c that sum up to eighteen, where c is twice the second integer (b), and the first integer (a) is six more than the second (b), we set up the following equations based on the given information:
- a = b + 6
- c = 2b
- a + b + c = 18
Substituting the expressions for a and c in terms of b into the third equation, we get:
(b + 6) + b + 2b = 18
Combining like terms, we have:
4b + 6 = 18
Subtracting 6 from both sides gives us:
4b = 12
Dividing both sides by 4, we find:
b = 3
Now we can find a and c:
- a = b + 6 = 3 + 6 = 9
- c = 2b = 2 × 3 = 6
Therefore, the three integers are 9 (a), 3 (b), and 6 (c).