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Question 824010: Three times the sum of the digits of a positive, two-digit integer is 50 greater than the difference of the digits. Reversing the digits decreases the number by 9. What is the number?
Found 2 solutions by lwsshak3, MathTherapy:
Answer by lwsshak3(11628)   (Show Source):
You can put this solution on YOUR website! Three times the sum of the digits of a positive, two-digit integer is 50 greater than the difference of the digits. Reversing the digits decreases the number by 9. What is the number?
***
let u=units digit
let t=tens digit
..
3(t+u)-(t-u)=50
(10t+u)-(10u+t)=9
..
3t+3u-t+u=50
10t+u-10u-t=9
..
2t+4u=50
9t-9u=9
..
18t+36u=450
36t-36u=36
add:
54t=486
t=9
4u=50-2t=50-18=32
u=8
What is the number? 98
Answer by MathTherapy(10552)  (Show Source):
You can put this solution on YOUR website!
Three times the sum of the digits of a positive, two-digit integer is 50 greater than the difference of the digits. Reversing the digits decreases the number by 9. What is the number?]
Number: 
You can do the check!!
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