SOLUTION: The sum of the digits of a two digit number is 10. The number formed by reversing the number is 4 less than 5 times the number. Find the original number? Can you please help me

Question 691413: The sum of the digits of a two digit number is 10. The number formed by reversing the number is 4 less than 5 times the number. Find the original number?
Can you please help me out with this question I really don't understand? thanks in advance
Found 2 solutions by josmiceli, stanbon:
Answer by josmiceli(19441) (Show Source): You can put this solution on YOUR website!
Let the tens digit =
Let the units digit =
-----------------------
The number looks like ab, like the number 36,
but the value of the number is ,
the same way that
-----------------------
given:
(1)
This is just talking about the digits and not the value
of the number
---------------
The VALUE of the number, as I said, is
and the value of the number formed by reversing the digits is

You are told that
(2)
You have 2 equations, (1) and (2), and 2 unknowns, so
it is solvable.
---------------
(2)
(2)
and
(1)
(1)
Substitute (1) into (2)
(2)
(2)
(2)
(2)
(2)
and, since
(1)
(1)
(1)
The number is 19
check answer:
(2)
(2)
(2)
(2)
(2)
OK

Answer by stanbon(75887) (Show Source): You can put this solution on YOUR website!
The sum of the digits of a two digit number is 10. The number formed by reversing the number is 4 less than 5 times the number. Find the original number?
---
Let the number be 10t+u ; t is the 10's digit; u is units digit
The reverse number is 10u + 5
--------
Equations:
t + u = 10
-------
10u+t = 5(10t+u)-4
===========
Simplify the bottom equation:
10u+t = 50t+5u-4
49t - 5u = 4
====
Substitute for t = 10-u for "t" and solve for "u":
49(10-u) - 5u = 4
490 - 54u = 4
54u = 486
u = 9
----
Now, solve for "t" using t + u = 10
---
t+9 = 10
t = 1
-----
The number is 10t+u = 10*1 + 9 = 19
------------
Cheers,
Stan H.
================
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