SOLUTION: When the digits of a two-digit number are reversed,the new number is 9 more than the original number,and the sum of digits of the original number is 11.What is the original? I re

Question 300399: When the digits of a two-digit number are reversed,the new number is 9 more than the original number,and the sum of digits of the original number is 11.What is the original?
I really need help with setting this up and how to check.
Please and Thank you!
Found 2 solutions by Fombitz, JBarnum:
Answer by Fombitz(32388) (Show Source): You can put this solution on YOUR website!
The original number is AB where the actual value is 10*A+B.
The reversed digits number is BA and its actual value is 10*B+A.
Nw turn the words into equations.
"sum of digits is 11"
1.
"the digits of a two-digit number are reversed,the new number is 9 more than the original number"

2.
Now you have two equations in A and B.
Add them together to eliminate A and solve for B. THen go back and solve for A.
To check your answer, run through the two requirements and make sure the digits meet them.
Answer by JBarnum(2146) (Show Source): You can put this solution on YOUR website!
original number: (ab)=10a+b
new number: (ba)=10b+a
a+b=11
(ba)=9+10a+b
hmm this is harder to explain then i thought.
[for example take number 45=(10(4)+5) if reversed its 54=(10(5)+4). 4+5=9 sum of digits of original number 45, (10(4)+5)+9=54 added 9 to the original 45 to get the new number 54.]
same type of thing for this problem
a and be have to be single whole numbers that add to get 11

________

________
check
(ab)=10a+b
(ab)=10(5)+(6)
(ab)=(56)
__
(ba)=10b+a
(ba)=10(6)+(5)
(ba)=(65)
__
(ba)=9+(ab)
(65)=9+(56)
(65)=(65)
CORRECT

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