Three times the sum of three consecutive integers is 72.What are the integers.?

Question

explain and make equation pleaseeeeeeee.

Answer 1

let x=first number....so every other number is based off of x....second number=x+1 and the third number=x+2.

the sum of three consecutive integers is then written as,
x + (x+1) + (x+2), which is the same as the first int. + sec int. + third int.

three times the sum is three times the whole thing...
3 [x + (x+1) + (x+2)] =72.

Solve for x.
1. Divide out 3 from both sides...
[x + (x+1) + (x+2)]=24
2. Combine like terms.
3x+3=24.
3. Divide both sides by 3. Divide the whole left side and the whole right side.
x+1=8.
4. x on one side, x=7.

Rember: "x" was the first integer so 7 is the first integer and the other two are 8 and 9.

Answer 2

3(x+1+x-1+x) = 72
three consecutive are x-1, x, and x+1
3x = 72
x = 24
23, 24, 25

Answer 3

Let the first integer be a.

Then we know that a + (a+1) + (a+2) = 72

This means that 3a +3 = 72
Then 3a = 69
And a = 23

So the integers are 23, 24, 25

Answer 4

too easy:

in my head: x + (x +1) + (x + 2) = 72->

3x + 3 = 72 -> 3x = 69-> x = 13.

23, 24, 25