Five times the sum of three consecutive integers is 150

9, 10, 11

Step-by-step explanation:

Note there is a difference of 1 between consecutive integers.

Let n be the smallest integer then n + 1 and n + 2 are the next 2 integers, then

5 (n + n + 1 + n + 2) = 150 (divide both sides by 5)

3n + 3 = 30 (subtract 3 from both sides)

3n = 27 (divide both sides by 3)

n = 9 and n + 1 = 9 + 1 = 10, n + 2 = 9 + 2 = 11

The 3 consecutive integers are 9, 10, 11

Step-by-step explanation:

Let the numbers be x, x + 1, x + 2

The sum will be = { x + x + 1 + x + 2}

= 3x + 3

And 5times the sum is 150

5 (3x + 3) = 150

Opening bracket

15x + 15 = 150

Collecting like terms

15x = 150 - 15

15x = 135

Dividing each term by 15

15x/15 = 135/15

x = 9

: the numbers are x, x + 1, x + 2

9, 9 + 1, 9 + 2

There are 9, 10 and 11