find three consecutive integers such that the sum between the first integer and twice the second integer is equal to ten less than the third integer answers.com

Answer 1

Step 1: We must represent the 3 consecutive integers with a combination of variables and constants.

Integer 1: n

Integer 2: n + 1

Integer 3: n + 2

Step 2: Devise an equation that meets the “such that” conditions.

The sum of the first integer and twice the second integer can be represented by

n + 2(n + 1)

Ten less than the third integer can be represented by

(n + 2) - 10

Setting these equal, we get

n + 2(n + 1) = (n + 2) - 10

Simplifying each side, we get

n + 2n + 2 = n - 8

3n + 2 = n - 8

2n = - 10

n = -5

The 3 integers are -5, -4, -3.

To prove this, plug the values back into the equation:

n + 2(n + 1) = (n + 2) - 10

-5 + 2(-4) = -3 - 10

-5 + (-8) = -13

-13 = -13