Posted by Astrid on Sunday, January 6, 2008 at 10:38pm.
I thought Damon did a pretty good job of explaining this question before
Ok, here is my approach, perhaps it will make sense to you.
Let the unit digit of the original number be x
Let the tens digit by y
then the original number was 10y+x
We were told the sum of the digits is 14, so
x+y=14, this is your first equation
the number reversed would be 10x+y
but this is 18 less than the original number, so....
10x+y + 18 = 10y+x , (since it was 18 less, I added 18 to make them "equal")
9x - 9y = -18
x-y = -2 , this is your second equation.
I will leave it up to you to solve them
I dont understand how to get the second equation
would you agree that according to my definition, the original number is 10y+x and the number reversed is 10x+y ???
your problem stated "the new number is 18 less than the original number." which translates into
10x+y < 10y+x by 18, so I added 18 to the "smaller" side to make them "equal", thus
10x+y + 18 = 10y+x
surely you can see how that simplifies to x-y=-2
So the answer would be 86?
Do the digits of 86 add up to 14?
Is 68 less than 86 by 18 ??
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