Question

Find three consecutive odd integers with the sum of 63

Answer 1

We have to set up a simple equation here. n+(n+2)+(n+4)=63. Now we solve. 3n+6=63, so 3n=57. n=19. n=19, n+2=21, and n+4=23. So the consecutive integers are, in order, 19, 21 and 23.

Answer 2

Answer:

19, 21, 23

Step-by-step explanation:

Let  x   be the first integer.

Let x  +  2  be the second integer.

Let  x  +  4  be the third integer.

The sum of the three consecutive odd integers is  63 , so your equation is:

(x) + (x+2) + (x+4) = 63

Then solve for x

(x)  + (x+2) + (x+4) = 63

x + x+2 + x+4 = 63

3x + 6 = 63

3x + 6 − 6 = 63  − 6

3x = 57

3x ÷ 3 = 57  ÷ 3

x = 19

Now that you know your first integer has a value of  

19 , substitute  x  =  19 into x + 2  and x +4

to find the values of the second and third integers.

x + 2                    x + 4

= (19 + 2)              = (19 + 4)

= 21                      = 23