Question 86887
I'm not sure how to take a story problem and put it into a equation. could you
help!!
When the digits of a two-digit number are reversed, the new number is 9 more
than the original number, and the sum of the digits of the original number is
11. What is the original number?
Help!!!!
How would you check it also?
Thanks so much!!!
 
<pre><b>
First notice that a two digit number like 74 is such that if you take the fist
digit 7, multiply it by 10, you get 70, and then if you add the second digit,
4,  to the 70, you get the number 74.   74 = 10×7 + 4

Another example: Notice that a two digit number like 29 is such that if you
take the fist digit 2, multiply it by 10, you get 20, and then if you add the
second digit 9 to the 20, you get the number 29.   28 = 10×2 + 9

Therefore, it's easy to see that 

any 2-digit number = 
           
          10 times its first digit + its second digit.
<font color = "blue" size = 5>
Let x = the first digit and y = the second digit.

Then the number = 10x + y and
the number reversed = 10y + x


>>...the sum of the digits of the original number 
is 11...<<

x + y = 11

>>...When the digits of a two-digit number are 
reversed, the new number is 9 more than the original 
number...<<

(10y + x) = (10x + y) + 9

simplifying:

  10y + x = 10x + y + 9

  9y - 9x = 9

Dividing thru by 9

    y - x = 1 

So you have this system if equations:

x + y = 11
y - x = 1

Solve that by substitution and get x=5, y=6

So the number is 56.

Checking in the words:
When the digits of 56 are reversed, you get the 
new number 65, and that is indeed 9 more than 56, 
and the sum of the digits of 56 is 5+6 which 
indeed does equal 11.

Edwin</pre>