Question 276733
when the digits of a two-digit number are reversed, the new number is 9 more than the original number, 
and the sum of the digits of the original number is 13. 
What is the original number?
:
Let x = the 10's digit
Let y = the units
then
10x + y = "the number"
:
Write an equation for each statement:
"the digits of a two-digit number are reversed, the new number is 9 more than the original number,"
10y + x = 10x + y + 9
Combine like terms
10y - y = 10x - x + 9
9y = 9x + 9
Simplify, divide by 9
y = (x + 1)
:
"the sum of the digits of the original number is 13."
x + y = 13
Substitute (x+1) for y
x + (x+1) = 13
2x = 13 - 1
2x = 12
x = 6 is the 10's digit
then since the sum is 13, obviously
y = 7 is the units
and
67 is the original two digit number
:
Check this in the statement:
"when the digits of a two-digit number are reversed, the new number is 9 more than the original number,
76 = 67 + 9
:
When the number 67 if reversed it is 9 more than the original