Question

The larger of two consecutive integers is 7 greater than twice the smaller. Find the integers. Find three consecutive odd integers such that the sum of the smallest and 7 times the largest is 68.

Answer 1

The two consecutive integers are -6 and -5. The three consecutive odd integers are 5, 7, and 9, where the smallest and 7 times the largest add up to 68.

Step-by-step explanation:

To solve the first part of the question, let's define the smaller integer as x. According to the problem, the larger integer is 7 greater than twice the smaller, so it can be represented as 2x + 7. Since the integers are consecutive, the larger one is also x + 1. Setting these two expressions for the larger integer equal to each other gives us an equation:

2x + 7 = x + 1.

Solving for x, we get x = -6. Therefore, the two consecutive integers are -6 and -5.

To find the three consecutive odd integers, let's designate the smallest odd integer as y. The next consecutive odd integers would be y + 2 and y + 4 respectively. The sum of the smallest and 7 times the largest, according to the problem, is 68:

y + 7(y + 4) = 68.

Expanding and solving for y, we get y = 5. Therefore, the three consecutive odd integers are 5, 7, and 9.