SOLUTION: Find three consecutive even integers such that the sum of twice the first and three times the third is fourteen more than four times the second.

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Question 1199434: Find three consecutive even integers such that the sum of twice the first and three times the third is fourteen more than four times the second.
Found 2 solutions by math_tutor2020, josgarithmetic:
Answer by math_tutor2020(3817) About Me  (Show Source):
You can put this solution on YOUR website!

A = 1st number
B = 2nd number
C = 3rd number
Each item is a consecutive even integer.
Eg: 4,6,8

A = x
B = x+2
C = x+4
where x is an even integer

The phrase "sum of twice the first and three times the third is fourteen more than four times the second" translates to
2A+3C = 4B+14

Apply substitution and solve for x.
2A+3C = 4B+14
2x+3(x+4) = 4(x+2)+14
2x+3x+12 = 4x+8+14
5x+12 = 4x+22
5x-4x = 22-12
x = 10

So,
A = x = 10
B = x+2 = 10+2 = 12
C = x+4 = 10+4 = 14

Check:
2A+3C = 4B+14
2*10+3*14 = 4*12+14
20+42 = 48+14
62 = 62
The answers are confirmed.

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Answer:
10,12,14

Answer by josgarithmetic(39620) About Me  (Show Source):
You can put this solution on YOUR website!
If they are n-2, n, n+2
then
2%28n-2%29%2B3%28n%2B2%29=14%2B4n
-
2n-4%2B3n%2B6=4n%2B14
n=14-2
n=12-------the middle of these integers.

10, 12, 14