Posted by **Astrid** on Sunday, January 6, 2008 at 10:38pm.

The sum of the digits of a two-digit number is 14. If the numbers are reversed, the new number is 18 less than the original number. Find the original number.

I know Ana asked this question, but i dont understand how to get the equations.

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**Reiny**, Sunday, January 6, 2008 at 11:01pm
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

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**Astrid**, Sunday, January 6, 2008 at 11:31pm
I dont understand how to get the second equation

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**Reiny**, Monday, January 7, 2008 at 12:04am
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

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**Astrid**, Monday, January 7, 2008 at 12:09am
So the answer would be 86?

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