SOLUTION: I'm 46 yrs old and am trying to refresh algebra. When the digits of a 2-digit number are reversed, the new number is 9 more than the original number, and the sum of the digits

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Question 459943: I'm 46 yrs old and am trying to refresh algebra.
When the digits of a 2-digit number are reversed, the new number is 9 more than the original number, and the sum of the digits of the original number is 11. What is the original number?

Found 2 solutions by MathLover1, richwmiller:
Answer by MathLover1(20850) About Me  (Show Source):
You can put this solution on YOUR website!

Tlet’s two -digits number be: xy
x+= the 10's digit
y+= units digit
The original number is: 10x+%2B+y
given:
the digits of a two-digit number are reversed, the new number is 9+more
than the original number
the sum of the digits of the original number is 11: x+%2B+y+=+11.....
you will have:
10y+%2B+x+=+10x+%2B+y+%2B+9
10y+-+y+=+10x+-+x+%2B+9
9y+=+9x+%2B+9.........both sides divide by 9
y+=+x+%2B+1

Replace y+with %28x%2B1%29 and plug it in x+%2B+y+=+11
x+%2B+x+%2B+1+=+11
2x+=+10
x+=+5......the 10's digit
then, the units digit will be
y+=+5+%2B+1
y+=+6
and the original number is: 10x+%2B+y=10%2A5%2B6=56
if the digits of a two-digit number are reversed, we will have a number 65

65-56=9...so, the new number is 9+more
than the original number

Answer by richwmiller(17219) About Me  (Show Source):
You can put this solution on YOUR website!
x+y=11
10x+y=10y+x+9
56 is the original number
new number is 65
check
65=56+9
6+5=11
ok