SOLUTION: the sum of the digits of a two-digit number is 10. If the digits are reversed, the number is increased by 72. What is the original number?

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Question 174725: the sum of the digits of a two-digit number is 10. If the digits are reversed, the number is increased by 72. What is the original number?
Found 2 solutions by Mathtut, josmiceli:
Answer by Mathtut(3670) About Me  (Show Source):
You can put this solution on YOUR website!
let the two digits be a and b
:
a+b=10..........eq 1
:
10b+a=10a+b+72--->9a-9b=-72..eq 2
:
re write eq 1 to a=10-b and plug it into eq 2
:
9(10-b)-9b=-72
:
90-9b-9b=-72
:
-18b=-162
:
highlight%28b=9%29
highlight%28a=10-9=1%29
:
original numberhighlight%28ab=19%29

Answer by josmiceli(19441) About Me  (Show Source):
You can put this solution on YOUR website!
Call the units digit n
Call the tens digit m
Given:
n+%2B+m+=+10
The actual value of the number is
10m+%2B+n
The value of the number with the digits reversed is
10n+%2B+m
Given:
10n+%2B+m+=+10m+%2B+n+%2B+72
(1) 9n+-+9m+=+72
(2) n+%2B+m+=+10
Multiply both sides of (2) by 9 and add to (1)
(1) 9n+-+9m+=+72
(2) 9n+%2B+9m+=+90
(3) 18n+=+162
n+=+9
n+%2B+m+=+10
m+=+10+-+9
m+=+1
The original number is 19
check:
91+=+19+%2B+72
91+=+91
OK