Find three consecutive even integers such that 3 times the sum of the first two integers is 12 less than 3 times the first integer

Question

Define a variable, write an equation, and solve the problem

 

Answer 1

Let n, n+2, and n+4 represent the three consecutive even integers.

3(n+n+2) = 3n-12

Solve for n by collecting variables in one side, constant terms in the other side,

3n = -18

n = -6

Answer: -6, -4, -2

Answer 2

Let x, y, and z represent the 3 consecutive even integers, respectively. 

We are given that 3 times the sum of the first two integers (i.e., 3(x + y)) is equal to 12 less than 3 times the first integer (i.e., 3x - 12). With this, we arrive at the following equation:

          3(x + y) = 3x - 12

Factoring out a 3 from both terms on the right hand side of the equation, we get the following:

          3(x + y) = 3(x - 4)

After we divide both sides of the equation by 3, we are left with the following:

          x + y = x - 4

Solve for y by subtracting x from both sides of the equation:

          y = -4

Now we need to find x (the even integer that precedes y) and z (the next even integer that follows y).

Since x is the even integer that precedes y, then ..... x = -6 

Since z is the next even integer following y, then ..... z = -2

Thus, the 3 consecutive even integers that satisfy the conditions stated in the problem are -6, -4, -2.