Find the smallest of three consecutive i there such that the sum of the first and second is 6 more than the third

Answer 1

Consecutive means in a row, so if your first number is x, then the second number is x + 1 as the second number is 1 larger than the initial number. The third consecutive number is x + 2.

According to the problem the first two consecutive numbers, x being the first number and (x + 1) as the second number added together or x + (x +1) is the same as 6 more than the third consecutive number (x+2) or (x + 2) + 6

This gives us:

x + (x + 1) = (x + 2) + 6

Simplifying gets: 2x + 1 = x + 8

Subtracting 1 from each side leaves us with 2x = x - 7.

We want all the x"s on one side, so we subtract x from each side and get x = 7.

Our consecutive numbers are 7, 8, and 9. The smallest of those numbers being 7.

Checking the answer the sum of the first two numbers 7 + 8 is 15 which is 6 more than the third number 9.

Answer 2

x+x+1 = x+2+6

2x + 1 = x + 8

x + 1 = 8

x=7

7,8,9, and 7+8 = 15 = 9+6