SOLUTION: please help me: the sum of the digits of a two-digit number is 12. if the digits are reversed, the new number is 15 more than twice the original number. what is the original num

Question 158041: please help me:
the sum of the digits of a two-digit number is 12. if the digits are reversed, the new number is 15 more than twice the original number. what is the original number?
Answer by jim_thompson5910(35256) (Show Source): You can put this solution on YOUR website!
Let t=tens digit and u=units digit

ANY two digit number can be written in the form of: tens_digit*10+units_digit

Algebraically this looks like: where "n" is any two digit number

If you reverse the digits of "n" you get where "m" is the reversed number (ie if n=27 then m=72)

Since the "sum of the digits of a two-digit number is 12", this means that

Also, since the "new number is 15 more than twice the original number", this tells us that

Distribute

Subtract from both sides. Subtract from both sides

Combine like terms.

Rearrange the terms.

So we have the system:

Multiply the both sides of the first equation by 19.

Distribute and multiply.

So we have the new system of equations:

Now add the equations together. You can do this by simply adding the two left sides and the two right sides separately like this:

Group like terms.

Combine like terms. Notice how the x terms cancel out.

Simplify.

Divide both sides by to isolate .

Reduce.

------------------------------------------------------------------

Now go back to the first equation.

Plug in .

Multiply.

Subtract from both sides.

Combine like terms on the right side.

Divide both sides by to isolate .

Reduce.

So our answer is and .

This means that the original number is and the new number is
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