algebraic word problem

Question

finnd three consecutive off positive integers such that two times the sum off all three integers is sixty five less than the product of the first and second integers

Answer 1

Let the three numbers be n-2, n, and n+2.  Their sum is 3n. Twice their sum is 6n.  The product of the first two is (n-2)n=n²-2n.  6n=n²-2n-65, and so n²-8n-65=0 or (n-13)(n+5)=0 or n=13, and the three integers are

11, 13, 15, or n=-5 and the three can't be positive.

 

Answer 2

2n+1 , 2n +3, 2n +5

 

@(2n +1 + 2n +3 + 2n +5 )= 12n +18 

 

 12n +18 =  ( 2n +1 ) ( 2n +3)- 65

 

 12n +18 = 4n^2 +8n -62

 

   4n^2 -4n -80 =0

 

     n^2 - n - 20 =0

 

     ( n - 5 ) ( n +4 ) =0

 

       n =5

 

     3 numbers : 11, 13 , 15 

Answer 3

Find three consecutive odd positive integers such that two times the sum of all three integers is sixty five less than the product of the first and second integers.

 

(2n+1)

(2n+3)

(2n+5)

 

2((2n+1)+(2n+3)+(2n+5))= (2n+1)(2n+3) - 65

 

2(6n+9) = 4n^2+8n+3 - 65

 

12n+18 = 4n^2+8n+3 - 65

 

4n^2 - 4n - 80 = 0

 

n^2 - n - 20 = 0

 

(n-5)(n+4) = 0

 

n = 5, n = - 4; but have to be positive integers

 

n = 5

 

(2n+1) = 11
(2n+3) = 13
(2n+5) = 15

 

check:

 

2(11+13+15) =? 11*13 - 65

 

2(39) =? 143 - 65

 

78 = 78 √