find three consecutive odd integer such that the sum of the first is two less than the second, and three more than the third is 70.
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Secondary School Math 5+3 pts
Find three consecutive odd integers such that the sum of the first is two less than the second and three more than the third is 70
Report by Kashishasiwal2006 01.09.2018
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Homosapiens45Ambitious
Given :The sum of the first, twice the second, and three times the third = 70
To find : The three consecutive odd integers.
Proof:
Let the first odd integer = x ,
The second consecutive odd integer = x+2 ,
The third consecutive odd integer = (x+2)+2 = x+4
So,The sum of the first, twice the second, and three times the third will be
x+2(x+2)+3(x+4)
Given the sum of the first, twice the second, and three times the third = 70
∴ x+2(x+2)+3(x+4) = 70
=> x+2x+4+3x+12 = 70
=> 6x+16 = 70
=> 6x = 70-16
=> 6x = 54
=> x = 54÷6
=> x = 9
Now we have;
first odd integer = x
=> x= 9
The second consecutive odd integer = x+2
=> x+2
=> 9+2 = 11
=> x = 11
The third consecutive odd integer = x+4
=> x+4
=> 9+4 = 13
=> x = 13
∴ the three consecutive odd integers x=9, x=11, x= 13.
Hence, the three consecutive odd integers are 9,11,and 13.
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Three consecutive odd integer so let's consider the following terms:
Let the first variable be x
let the second variable be x + 2
third variable ( x + 2) + 2 = x + 4
So sum of first less than second and three more than third so we will apply following methods
=> x + (x + 2 - 2 ) two less
=> + ( x + 4 + 3 ) three more than third
So bringing the equation together
=> x + ( x + 2 - 2 ) + ( x + 4 + 3) = 70
=> x + x + x + 4 + 3 = 70
=> 3x + 7 = 70
=> 3x = 63
=> x = 21 first odd integer
=> x + 2 = 21 + 2 = 23 second odd integer
=> x + 4 = 21 + 4 = 25 third odd integer