Find three consecutive odd integers such that the sum of the first number,two less than the second number and three more than the third number is 70.
Answers
Step-by-step explanation:
Let the 3 consecutive odd integer be ;x,x+2,x+4
x+2(x+2)+3(x+4)=70
x+2x+4+3x+12=70
6x+16=70
6x=54
x=9
∴3 no : are : 9,11,13
Given:
✰ They are three consecutive odd integers.
✰ The sum of the first number is two less than the second number and three more than the third number is 70.
To find:
✠ The three consecutive odd integers.
Solution:
Let the first odd integer be x.
Let the second odd integer be x + 2 [ We have not taken x + 1 because they are 'odd' integers not even, taking 1 the number will become even, infact 1 is odd but the whole number is x + 1 adding it by x makes it even ].
Let the third number be ( x + 2 ) + 2
= x + 4
We are given that the sum of the first number is two less than the second number that means 2 will be substracted from second number and first number will be added ( sum ) to it.
➛ x + (x + 2 - 2 )
and three more than the third number, we will add three to third number.
➛ + ( x + 4 + 3 )
Together we have, there result is 70
➤ x + (x + 2 - 2 ) + ( x + 4 + 3 ) = 70
➤ x + x + x + 4 + 3 = 70
➤ 3x + 7 = 70
➤ 3x = 70 - 7
➤ 3x = 63
➤ x = 63/3
➤ x = 21
∴ The first odd integer number = 21
Now,
➤ The second odd integer = x + 2
➤ The second odd integer = 21 + 2
➤ The second odd integer = 23
∴ The second odd integer number = 23
➤ The third odd integer number = x + 4
➤ The third odd integer number = 23 + 4
➤ The third odd integer number = 25
∴ The third odd integer number = 25