SOLUTION: A number of two digits are equal to 6 times the sum of the digits, and the number formed by reversing the digits exceeds four times the sum of the digits by 9. what are the origina

Question 698705: A number of two digits are equal to 6 times the sum of the digits, and the number formed by reversing the digits exceeds four times the sum of the digits by 9. what are the original numbers?

(I have worked out the numbers to be 5 and 4, but this is through guess and check. I don't know how to write out the equation)
THANK YOU!
Found 2 solutions by josmiceli, stanbon:
Answer by josmiceli(19441) (Show Source): You can put this solution on YOUR website!
Let the 10s digit =
Let the units digit =
Note that the actual value of the number is

---------------
given:
(1)
(2)
-----------------------------
(1)
(1)
(1)
(1)
and
(2)
(2)
(2)
(2)
By substitution:
(2)
(2)
(2)
(2)
and
(1)
(1)
(1)
The original number is 54
Whatever you did was pretty good.

Answer by stanbon(75887) (Show Source): You can put this solution on YOUR website!
A number of two digits are equal to 6 times the sum of the digits, and the number formed by reversing the digits exceeds four times the sum of the digits by 9. what are the original numbers?
----
Let the number be 10t+u ( t is the ten's digit; u is the unit's digit)
-----
Equation:
10t+u = 6(t+u)
10u+t = 4(t+u) + 9
------
Rearrange:
4t -5u = 0
-3t+6u = 9
-----------------
Modify:
12t - 15u = 0
-12t +24u = 36
---
Add and solve for "u":
9u = 36
u = 4
-----
Solve for "t":
12t = 15u
t = 15*4/12
t = 5
-----
Ans: The number is 10t+u = 54
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Cheers,
Stan H.
====================
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