Find two consecutive odd integers such that their product is 107 more than 6 times their sum.

Question

please answer this question for me

Answer 1

 

Let     N = 1st number

Let N+2 = 2nd number ... the next odd number is 2 more than the first

 

   Product is the result of multiplication.  N(N+2) 

   Sum is the result of adding:  N+N+2 = 2N+2

   6 times the sum:  6(2N+2) = 12N + 12

   107 more:  12N+12+107 = 12N + 119

 

Our equations is:

 

   Product is 107 more than 6 times their sum

   N(N+2) = 12N+119

   N² + 2N = 12N +119

   N² + -10N - 119 = 0

   (N - 17)(N + 7) = 0

 

Using 0 product property,

 

   N-17 = 0 ... N = 17

   Numbers are 17 and 19

 

   N+7 = 0 ... N = -7

   Numbers are -7 and -5

 

CHECK: (17 and 19)

     Product:  17(19) = 323

     Sum:  17+19 = 36

     6 times the sum:  36(6) = 216

 

     323 - 107 = 216 

 

CHECK:  (-7 and -5)

     Product:  -7(-5) = 35

     Sum:  -7+-5 = -12

     6 times the sum:  6(-12) = -72

 

     35 - 107 = -72

 

Answer 2

Follow the procedure I presented in your previous post on the same type of problem.

To learn how to do a problem of this type you must make some attempt.