find three consecutive integers such that the sum of the first and the third is sixteen more than the second

Answer 1

Let's let x represent the first number or the smallest one. Then the second number would be represented with x + 1 and the third would be represented with x + 2.

The sum of the first and the third would be written as: x + x+2

Sixteen more than the second would be written as: x+1 +16

These two expressions are equal to one another: x + x + 2 = x + 1 + 16

Now we can solve for x, the first number:

x + x + 2 = x + 1 + 16

2x + 2 = x + 17

2x + 2 - x = x + 17 - x

x + 2 = 17

x + 2 -2 = 17 - 2

x=15

The first number would be 15, the second 16, and the third 17.