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Question 557603: A two-digit number is eight times the sum of its digit. When the number is added to the number obtained by reversing the digits, the sum is 99. Find the original number.
Thank you.
Answer by ankor@dixie-net.com(22740)   (Show Source):
You can put this solution on YOUR website! Let x = the 10's digit
Let y = the units
then
10x+y = the two-digit number
:
A two-digit number is eight times the sum of its digit.
10x + y = 8(x + y)
10x + y = 8x + 8y
10x - 8x + y - 8y = 0
2x - 7y = 0
:
When the number is added to the number obtained by reversing the digits, the sum is 99. Find the original number.
10x+y + 10y+x = 99
11x + 11y = 99
simplify, divide by 11
x + y = 9
Multiply this equation by 2, subtract the 1st equation
2x + 2y = 18
2x - 7y = 0
-------------subtraction eliminates x, find y
9y = 18
y = 2
then
x + 2 = 9
x = 7
:
72 is the number, you can check this in both statement equations:
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