Question

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

Answer 1

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.

Answer 2

Hello mate here is your answer

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

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