Question

Find two consecutive odd numbers such that the difference

between twice the greater number and one-third of the
smaller number is 29.

Answer 1

15,17

Explanation:

Let the two consecutive odd numbers be (2x + 1) and 2x + 3

Twice the greater number : 2* (2x + 3) = 2*2x + 2*3

= 4x + 6

`\sf \text{One third of smaller number=$(1)/(3)*(2x + 1)$}\\`

`\sf = (1)/(3)*2x + (1)/(3)*1\\\\=(2x)/(3)+(1)/(3)`

Difference = 29

`\sf 4x + 6 -( (2x)/(3) + (1)/(3))=29\\\\ 4x + 6 -(2x)/(3)-(1)/(3)\\\\\text{\bf Multiply the entire equation by 3}\\\\3*4x + 3*6 - 3*(2x)/(3)-3*(1)/(3)=3*29\\\\ 12x + 18 - 2x - 1 = 87`

Combine the like terms.

12x - 2x + 18 - 1 = 87

10x + 17 = 87

Subtract 17 from both sides

10x = 87 - 17

10x = 70

Divide both sides by 10

x = 70/10

x = 7

2x + 1 = 2*7 + 1

= 14 + 1

= 15

2x + 3 = 2*7 + 3

= 14 + 3

= 17

`\sf \boxed{\text{\bf The consecutive odd numbers are 15,17}}`