Math Equation

[Tavrobel]

the sum of three consecutive positive even intergers is equal to two more than seven times half the number. What are the three even positive intergers?

To me I'm imagining it looks like this.

a + b + c = 2 + 7 x 1/2n what's a,b, and c?

I guess it's a bit late, but here it is.

So, the problem is telling us that we have three consecutive, positive, even integers. This is all important information, because it tells us that:
A) our integers are positive; negative answers will be incorrect
B) they are consecutive, which means that each variable can be stated in terms of the others
C) even, which means that they're even
D) the ambiguous "number" is the same on both sides

Three unknowns, more than three equations. Our equation that we get from the problem statement looks like:
a + b + c = (7/2)*x + 2
... solving for x, which is what your initial guess was.

But we know from our statements above, that each of the integers can be stated in terms of the others, which means that b = something*a. c = something*b; therefore, even c can be expressed in terms of a. We know that they're even, which means that each number is offset by 2 in sequence. If they were regular consecutive integers, we could just add +1, instead.

a (our first integer)
b = (a + 2)
c = (b + 2)

Substitute in what we know about b, into the equation for c, so that we can get rid of variables.
c = ((a + 2) + 2)

So, our integers now look like a, (a + 2), and (a + 4). Put those back into the problem statement's equation:
a + (a + 2) + (a + 4) = (7/2)x + 2
3a + 6 = (7/2)x + 2
3a + 4 = (7/2)x
6a + 8 = 7x

If x is a given, then you could solve the problem to get a number for a. Then add 2 to a to get b, and add 4 to get c. Otherwise, you have two unknowns, and only one equation: you can't solve the problem. However, we've been told that it's the same number on both sides; therefore, x = a.

6a + 8 = 7a
8 = (7a - 6a)
a = 8

6*8 + 8 = 7*8
48 + 8 = 56?
a = 8, b = 10, c = 12

QED