find the three consecutive even integers such that three times the first equals the sum of the other two
let the 3 numbers be
x, x+1, and x+2
3x = x+1 + x+2
3x = 2x+3
x = 3
Finish it up, and check my answer to see if it meets the given conditions
To find the three consecutive even integers, let's use algebra to solve the problem.
Let's assume the first even integer as x. Since they are consecutive even integers, we can represent the second even integer as x + 2 (since any even integer plus 2 gives another even integer), and the third even integer as x + 4.
According to the problem, three times the first even integer (3x) is equal to the sum of the other two. So, we can write the equation:
3x = (x + 2) + (x + 4)
Now, let's solve the equation:
3x = x + 2 + x + 4
3x = 2x + 6
3x - 2x = 6
x = 6
Now that we have the value of x, which is 6, we can find the three consecutive even integers:
First even integer: x = 6
Second even integer: x + 2 = 6 + 2 = 8
Third even integer: x + 4 = 6 + 4 = 10
Therefore, the three consecutive even integers that satisfy the given condition are 6, 8, and 10.