The Sum of three consecutive odd integers is 45?????

x

x+2

x+4

3x + 6 = 45

3x = 39

x = 13

so 13, 15, 17 are the 3 consecutive odd integers

If n is the first of these odd integers, then the next odd integer is n+2 and the one after that is n+4. If their sum is 45, then:

n + (n+2) + (n+4) = 45

You can simplify and solve this for n. Then once you have n, you can find out what n+2 and n+4 are to get all three numbers:

3n + 6 = 45

3n = 39

n = 13, so the other numbers are 15 and 17

Assuming the first odd integer is x

then

x + x+2 + x+4=45

3x + 6=45

3x=39

x=13

Integers are 13, 15, 17

Let x = 1st odd integer

Let (x + 2) = 2nd odd integer

Let (x + 4) = 3rd odd integer

Add them all and let the total be 45

x + (x+2) + (x+4) = 45

x + x + 2 + x + 4 = 45

3x + 6 = 45

Then start transposing:

3x = 45 - 6

3x = 39

x = 39 / 3

And voila! Here is your answer

x = 13

x + 2 = 15

x + 4 = 17

13, 15, 17 in actuality you divide 40 5 through 3 it incredibly is 15 and to locate 3 consecutive unusual integers in simple terms use the unusual integer suitable in the previous it (13) and the only suitable after it (17) this 13, 15, 17

x + x + 2 + x + 4 = 45

3x + 6 = 45

3x + 6 - 6 + 45 - 6

3x = 39

3x / 3 = 39 / 3

x = 39/3

x = 13

- - - - - - - --

The first integer is 13

The second integer is 15

The third integer is 17

- - - - - - - - -s-

45 / 3 = 15

answer = 13, 15 , 17

x+(x+2)+(x+4)=45 where x is odd

3x+6=45

x=13

the numbers 13 15 and 17

let the integers be x-2,x,x+2,-----add them up 3x=45

x=15....hence the numbers are 13,15,17.....bye