# 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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Let the three even integers be `x, x+2, x+4` (with 2 difference)

According to the question, two times first i.e. `2x` , 3 times the third i.e.

`3(x+4) ` and 4 times the second i.e. `4(x+2)```

`2x + 3(x+4) = 14 + 4(x+2)`

` `

`2x + 3x + 12 = 14 + 4x + 8`

separate x terms and constants

`2x + 3x - 4x = 14 + 8 - 12`

`x = 10`

`therefore` the required numbers are 10, 12 and 14

lets verify the answer:

`10 + 3*14 = 14 + 4*12`

`52 = 52`

Hello!

Any even integer has the form 2*n for some integer n. The next even integer is 2n+2 and the next is 2n+4.

It is given that

2*(2n) + 3*(2n+4) = 4*(2n+2) + 14.

Gather terms with and without n and obtain

10n+12=8n+22,

2n=10,

n=5.

So the first even integer is 10, the second is 12 and the third is 14.