SOLUTION: I'm doing a digit problem and i don't understand. The problem is: the sum of the digits of a two digit number is 10. if the digits are reversed, the new number is 1 less than twice

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Question 36145: I'm doing a digit problem and i don't understand. The problem is: the sum of the digits of a two digit number is 10. if the digits are reversed, the new number is 1 less than twice the original number. what is the original number.
we supposed to use 10T+U for the original number and 10U+T for the reversed number but I don't understand how to set the 2 equations up- the one with the original and reversed number then the one with the sum of the digits of a two digit number is 10. Please help me-thank-you
Answer by Paul(988) About Me  (Show Source):
You can put this solution on YOUR website!
Let the 10-digit be x
Let the one digit be y
x+y=10
y=10-x (subsitution)
The original number is 10x+y
Equation:
Reversed digits -->10y+x

10y+x=2(10x+y)-1

10y+x=20x+2y-1

Subsitute for y:

10(10-x)+x=20x+2(10-x)-1

Expand:

100-10x+x=20x+20-2x-1

Simplfy:

100-9x=18x+19

27x=81

x=3

y=10-3

y=7

Hence, the 10-digit number is 3 and the one digit number is 7.  The number is 37.

Paul.