Question 300947: 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 11.What is the original number? (No one seems to help answer my question) PLEASE someone help me! Much Thanks to whoever does =)
Answer by ankor@dixie-net.com(22740)   (Show Source):
You can put this solution on YOUR website! 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 11.
What is the original number?
:
Let x = the 10's digit
Let y = the units digit
:
10x + y = the original number
and
10y + x = the reversed number
:
The equation for the 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 y's on the left and x's on the right
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 11."
x + y = 11
From the 1st equation substitute (x+1) for y
x + (x+1) = 11
2x = 11 - 1
2x = 10
x= 10/2
x = 5
then
y = 5 + 1
y = 6
:
What is the original number? 56 is the original number
:
:
See if that satisfies the 1st statement:
"the digits of a two-digit number are reversed,the new number is 9 more than the original number,
65 = 56 + 9; confirms our solution
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