SOLUTION: I am thinking of three consecutive positive numbers. If i multiply the first with the third and then add the second, the result is 41. Let x be the smallest number.

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Question 99076: I am thinking of three consecutive positive numbers. If i multiply the first with the third and then add the second, the result is 41. Let x be the smallest number.
Answer by rmromero(383) About Me  (Show Source):
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I am thinking of three consecutive positive numbers.  If i multiply the first with the third and then add the second, the result is 41.  Let x be the smallest number.



What is asked in the problem?

  find the 3 consecutive positive numbers.



Given:

  the first is multiplied with the third and then add the second, the result is 41.



Representation:

  Let x be the smallest number

     x + 1 be the second number

    x + 2 be the third/greatest number



Equation:



          x (x+2) + x+1 = 41

       x^2 + 2x + x + 1 = 41

           x^2 + 3x + 1 = 41

          x^2 + 3x - 40 = 0

        (x + 8) (x - 5) = 0



   x + 8 = 0    and    x - 5 = 0

     x = -8            x = 5



Since we are asked to find the three consecutive POSITIVE number

we reject -8.



x = 5

x + 1 = 5 + 1 = 6

x + 2 = 5 + 2 = 7



Therefore the three consecutive POSITIVE integers are 5, 6, and 7.