Question

find three consecutive numbers such that three times the first number is equal to 9 more than the sum of the other two.

Answer 1

Answer:

12, 13, 14

Step-by-step explanation:

Three consecutive numbers means three numbers in a row, so if n is the first number, the second number is one more than  n (which is n+1) and the third number is one more than that (which is n+2)

So three times (3 x) the first number (n) is equal to (=) nine more than (9 +) the sum of the other two ( (n+1) + (n+2) )

3 times n can be written as 3n

So 3n = 9 + (n+1) + (n+2)

3n = 9 + n + 1 + n + 2

3n = 9 + 1 + 2 + n + n

3n = 12 + 2n

Subtract 2n from each side

3n - 2n = 12 + 2n - 2n

1n = 12

n = 12

So the first number is 12, the second number is 12 + 1 (13) and the third number is 12 + 2 (14)

The numbers are 12, 13, 14

Hope this helps :)