SOLUTION: Five times the sum of the digits of a two-digit number is 13 less than the original number. If you reverse the digits in the two-digit number, four times the sum of its two digits

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Question 957727: Five times the sum of the digits of a two-digit number is 13 less than the original number. If you reverse the digits in the two-digit number, four times the sum of its two digits is 21 less than the reversed two-digit number.
The difference of the original two-digit number and the number with reversed digits is

Answer by josmiceli(19441) About Me  (Show Source):
You can put this solution on YOUR website!
Let +a+ = the tens digit
Let +b+ = the units digit
------------------------
(1) +5%2A%28+a+%2B+b+%29+=+10a+%2B+b+-+13+
(2) +4%2A%28+b+%2B+a+%29+=+10b+%2B+a+-+21+
--------------------------------
(1) +5a+%2B+5b+=+10a+%2B+b+-+13+
(1) +5a+-+4b+=+13+
---------------------
(2) +4b+%2B+4a++=+10b+%2B+a+-+21+
(2) +-3a+%2B+6b+=+21+
----------------------
Multiply both sides of (2) by +5+
and both sides of (1) by +3+
and add the equations
(1) +15a+-+12b+=+39+
(2) +-15a+%2B+30b+=+105+
-------------------------
+18b+=+144+
+b+=+8+
and
(1) +5a+-+4%2A8+=+13+
(1) +5a+=+13+%2B+32+
(1) +5a+=+45+
(1) +a+=+9+
The original number is 98
Number with reversed digits is 89
+98+-+89+=+9+
-----------------
check:
(1) +5%2A%28+a+%2B+b+%29+=+10a+%2B+b+-+13+
(1) +5%2A%28+9+%2B+8+%29+=+10%2A9+%2B+8+-+13+
(1) +5%2A17+=+98+-+13+
(1) +85+=+85+
OK
(2) +4%2A%28+b+%2B+a+%29+=+10b+%2B+a+-+21+
(2) +4%2A%28+8+%2B+9+%29+=+10%2A8+%2B+9+-+21+
(2) +4%2A17+=+89+-+21+
(2) +68+=+68+
OK