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

Answer:

15,17

Step-by-step 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

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

                                            $\sf = \dfrac{1}{3}*2x + \dfrac{1}{3}*1\\\\=\dfrac{2x}{3}+\dfrac{1}{3}$

Difference = 29

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

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

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