Expert: Pawan Musale - 9/7/2009

Question
QUESTION: Can You help me solve this equation... I really need you to show me the procedure rather than just give me the answer please... I need it ASAP the product of two odd consecutive integers is 38 less than the square of the second integer. what are the two numbers???

ANSWER: Martha,

This is pretty simple. Please find the attached scanned document <Two Integers.jpg> for the answer.

I hope this will solve your problem. Please let me know if you need additional information.

All the best.

Regards,
Pawan

---------- FOLLOW-UP ----------

QUESTION: thanks a lot Pawan... I have another question: find three consecutive odd integers such that the product of the first and third, minus the product of the first and second is five times more than the third... This is what I did... (x)(x+4)-[(x)(x+2)]=5(x+4)
      x^2+4x-x^2-2x=5x+20
      2x=5x+20
      -3x=20
       x=-21/3
I know this is not the answer but i just dont know what it is that I am doing wrong... could you please help me!!

Answer
Martha,

The question itself is wrong somewhere. The term '5 time more than' doesnot make any sense here. I believe that it should be only '5 more than third'.

If we assume so, your equation will become

(x)(x+4)-[(x)(x+2)]=5 +(x+4)
2x = 9 + x
x = 9

Hence two numbers are 9, 11, 13.

I hope this will help.

Let me know if you need additional information.

All the best.

Regards,
Pawan