SOLUTION: The sum of 3 numbers is 78, the third 4 more than the second, the first is two less than twice the second

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Question 1065513: The sum of 3 numbers is 78, the third 4 more than the second, the first is two less than twice the second
Found 2 solutions by Edwin McCravy, Alan3354:
Answer by Edwin McCravy(20060) About Me  (Show Source):
You can put this solution on YOUR website!
The sum of 3 numbers is 78,
Let the 1st number = x
Let the 2nd number = y
Let the 3rd number = z

x + y + z = 78

the third 4 more than the second,
z = y + 4

the first is two less than twice the second
x = 2y - 2

Substitute (2y - 2) for x, and (y + 4) for z in x + y + z = 78

(2y - 2) + y + (y + 4) = 78

Solve that for y, then substitute what you get for y in

x = 2y - 2 to find x, and in z = y + 4 to find z.

If you have trouble, ask me in the thank-you note form below
and I'll get back to you by email.  No charge ever.

Edwin

Answer by Alan3354(69443) About Me  (Show Source):
You can put this solution on YOUR website!
The sum of 3 numbers is 78, the third 4 more than the second, the first is two less than twice the second
n = 2nd number
---
n + n+4 + 2n-2 = 78