SOLUTION: he sum of a three-digit number is 15. If the digits are reversed and the resulting number is added to the original the sum is 1029. if the resulting number is subtracted from the o

Algebra ->  Customizable Word Problem Solvers  -> Numbers -> SOLUTION: he sum of a three-digit number is 15. If the digits are reversed and the resulting number is added to the original the sum is 1029. if the resulting number is subtracted from the o      Log On

Ad: Over 600 Algebra Word Problems at edhelper.com


    

Question 327350: he sum of a three-digit number is 15. If the digits are reversed and the resulting number is added to the original the sum is 1029. if the resulting number is subtracted from the original the difference is 693. what is the original number?
Answer by jvan(28) About Me  (Show Source):
You can put this solution on YOUR website!
Thanks for re-asking this question ^.^
Let the three digit number be represented by the variables xyz, with x being the values in the hundreds place, y being the number in the tens place, and z being the number in the ones place. Rewriting the problem into equation form you get the following. "The sum of a three-digit number is 15": x%2By%2Bz=15. "If the digits are reversed and the resulting number is added to the original the sum is 1029": %28xyz%29%2B%28zyx%29=1029. "The resulting number is subtracted from the original the difference is 693": %28xyz%29-%28zyx%29=693. (Note: it's great we have three equations because we have three unknown variables. If we have more unknown variables than equations, we won't be able to solve for the variables =)) Okay, let's solve for the variables. You can add the last two equations together to get: %28xyz%29%2B%28zyx%29+%2B+xyz%29-%28zyx%29=1029%2B693. You get 2xyz=1722. Divide both sides by 2 to get xyz=861. Now you plug this back in and solve for zyx to get zyx=168 =]
Both of these numbers are reverse of one another and if you add up their three digits you get 15 as a sum! Thus, the three digit number is 861.
****---->Please visit my site at http://myonlinetutor.webs.com and become a member \(^o^)/! Thanks!