Final answer:
By setting up an equation to represent the sum of twice the smallest even integer and three times the largest, we find that the three consecutive even integers that satisfy the given condition are 6, 8, and 10.
Step-by-step explanation:
To find three consecutive even integers where the sum of twice the smallest number and three times the largest is 42, we will set up an equation. Let's denote the smallest even integer as x, the middle integer as x+2, and the largest integer as x+4 (since even numbers are 2 units apart). The equation based on the problem statement becomes: 2x + 3(x + 4) = 42. Simplifying this, we get 2x + 3x + 12 = 42, which simplifies further to 5x + 12 = 42. Subtracting 12 from both sides gives us 5x = 30, and dividing both sides by 5 gives us x = 6. Therefore, our three consecutive even integers are 6, 8, and 10 (i.e. x, x+2, x+4).