Find three consecutive even integers so that the twice the sum of the second and third is twelve less than six times the first.
Let the numbers be x,x+2, x+4
Then we have
2(x+2 + x+4) = 6x-12
Now just find x
To solve this problem, let's start by assuming that the first even integer is x.
The second even integer would then be (x + 2) since consecutive even integers are always two units apart.
The third even integer would be (x + 4) since it's also two units apart from the second even integer.
According to the problem statement, twice the sum of the second and third integers is twelve less than six times the first. Mathematically, this can be represented as:
2((x + 2) + (x + 4)) = 6x - 12
Simplifying the equation, we can distribute the 2 to the sum of the second and third integers:
2(x + 2 + x + 4) = 6x - 12
Now we can simplify further by combining like terms:
2(2x + 6) = 6x - 12
Next, distribute the 2 to each term inside the parentheses:
4x + 12 = 6x - 12
To isolate the variable x, let's move the constant terms to the opposite side of the equation:
4x - 6x = -12 - 12
Simplifying further, we have:
-2x = -24
To solve for x, let's divide both sides of the equation by -2:
x = -24 / -2
Simplifying the division gives us:
x = 12
So, the first even integer is x = 12.
Now, let's find the second and third integers by substituting the value of x into our earlier expressions:
Second even integer: (x + 2) = 12 + 2 = 14
Third even integer: (x + 4) = 12 + 4 = 16
Therefore, the three consecutive even integers that satisfy the given condition are 12, 14, and 16.