11.6k views
4 votes
three consecutive even integers such that the sum of the smallest and twice the second is 8 more than the third. what are the numbers?

User Squiguy
by
8.4k points

1 Answer

4 votes

Final answer:

To find the three consecutive even integers, let's represent them as x, x+2, and x+4. Then, solve the equation that represents the sum of the smallest and twice the second being 8 more than the third. The solution is x = 4, so the three consecutive even integers are 4, 6, and 8.

Step-by-step explanation:

To solve this problem, let's represent the three consecutive even integers as x, x+2, and x+4.

The sum of the smallest integer and twice the second integer can be represented as x + 2(x+2).

And the third integer is x+4.

According to the problem, the sum of the smallest and twice the second is 8 more than the third:

x + 2(x+2) = (x+4) + 8.

Simplifying this equation, we get: x + 2x + 4 = x + 12.

Combining like terms, 3x + 4 = x + 12.

Subtracting x from both sides, we have: 2x + 4 = 12.

Subtracting 4 from both sides, we get: 2x = 8. Finally, dividing both sides by 2, we find x = 4.

Therefore, the three consecutive even integers are 4, 6, and 8.

User Arjun J Rao
by
8.1k points