Ok, I'm getting ready to go to school this summer, and in the meantime, I'm trying to learn as much as I can in regards to math (since physics is going to be my major). At this point, I've been working on a book that goes over math word problems, and I have gotten hung-up on just one problem in the book that I can't figure out.

Here is the word problem:

"In a two-digit number, the ten's digit is 3 more than the one's digit. If the digit's are reversed, the difference between the two numbers is 27. Find the number."

Now, here's how I went about trying to solve this:

x = first number (one's digit)

x+3 = second number (ten's digit)

Then, to account for placing the first and second numbers in the tenth's place for regular and reverse, I tried this equation:

10(x+3)+x=10x+x+3+27

The first equation before the equals is the original number, the second one is reversing the numbers, then I added 27 to compensate for the difference that is supposed to be between them. However, when I set the equation up like this, all the "x"s end up cancelling themselves out, like so.

10x+30+x=11x+30

11x+30=11x+30

Hence, I end up cancelling it all out. Since the book I am studying from doesn't explain the exact formula when you are using the "difference" between the two numbers, I'm not sure how else to do this.

I also tried this equation, but I still end up cancelling out the x's:

10(x+3)+x-(10x+x+3)=27

10x+30+x-10x-x-3=27

11x+30-11x-3=27

11x-11x+27=27

Can someone please tell me what I'm doing wrong here and the correct way to solve for "x", hence solving for the two digits?