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

Question

Find the integers. helping my niece with homework what is this????

how do you get the answer is what I really need to know thanks!!!

Answer 1

x
x+2
x+4

3x + 6 = 45

3x = 39

x = 13

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

Answer 2

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

Answer 3

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

Answer 4

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

Answer 5

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

Answer 6

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

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The first integer is 13

The second integer is 15

The third integer is 17

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Answer 7

45 / 3 = 15

answer = 13, 15 , 17

Answer 8

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

3x+6=45
x=13

the numbers 13 15 and 17

Answer 9

let the integers be x-2,x,x+2,-----add them up 3x=45
x=15....hence the numbers are 13,15,17.....bye