Find the sum of three consecutive odd integers if the sum of the first two integers is equal to twenty-four less than four times the third integer

Question

It says to write an equation and solve each problem. I need help translating please. 

Answer 1

An odd integer can be expressed as x = 2n+1 where n is an integer. (Therefore making 2n an even number and 2n+1 an odd number).

Three consecutive odd integers can then be expressed as  (2n+1), (2n+3), and (2n+5).

The sum of the first two integers is (2n+1) + (2n+3) = 4n+4 = 4(n+1). It is supposed to be 24 less than 4 times the third integer, or

4(2n+5)-24.

Setting both equal and solving for n:

4(n+1) = 4(2n+5)-24

 

24 = 4·6, so you can factor out 4 in the right-hand expression.

 

4(n+1) = 4(2n+5-6)   or   4(n+1) = 4(2n-1)

 

Cancel out factor 4 and solve for n.

 

n+1 = 2n-1      this yields   n=2

 

Now plug n into your three consecutive odd integer expressions:  (2n+1) = 5, (2n+3) = 7, (2n+5) = 9. The sum is therefore 5+7+9 = 21.

 

Double check your answer:  

Sum of first two integers:  5+7 = 12

24 less than 4 times third integer:  4·9 - 24 = 36-24 = 12.

Answer 2

The three consecutive odd integers can be written as: x, x+2, x+4. 

 

The sum of the first two integers: (x) + (x+2)

 

Is: =

 

Twenty four less than four times the third 

integer: 24 - 4*(x+4)

 

(x) + (x+2) = 24 - 4*(x+4)

x+x+2 = 24-4x-16

2x+2 = 8-4x

6x+2 = 8

6x = 6

x = 1.

 

Therefore,  the three consecutive numbers are: 1, 3, 5; and the sum of these three numbers are: 1+3+5=9.