Question

The sum of the digits of a certain two-digit number is 7. Reversing its digits increases the number by 9. What is the number?

Answer 1

Let's say 'x' is the first digit in the number, and 'y' is the second one.
You already know that x + y = 7 .  Now here comes the fun part:

'x' is in the ten's place, so its value in the number is 10x , and the value
of the whole number is (10x + y).

If you flip the digits around, then 'y' is in the ten's place, its value is 10y ,
and the value of the whole new number becomes (10y + x) .

The problem tells you that when they're flipped around, the value is 9 more.

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

10y + x = 10x + y + 9

Just to make it neater and easier to handle, let's subtract 1x and 1y from each side:

9y = 9x + 9

Now.  What to do with this.
Remember that one up at the top ... x + y = 7 ?  That's just what we need now.
Call it [ y = 7 - x ], and we can plug that into the one we're struggling with:

9y = 9x + 9

9(7 - x) = 9x + 9

63 - 9x = 9x + 9

Subtract 9 from each side:  54 - 9x = 9x

Add 9x to each side:    54 = 18x

Divide each side by 18:    3 = x

Great.    y = 7 - x      y = 4

The original number is  34 .    3 + 4 = 7
Flip the digits, and you have 43 .
43 is 9 more than 34 .
yay.
======================================

That was the elegant but tedious way to do it.
Here is the brute-force but easy way to do it:

List of all the 2-digit numbers whose digits add up to 7, and their flips :
16 . . . . . 61
25 . . . . . 52
34 . . . . . 43
43 . . . . . 34
52 . . . . . 25

Do you see a number that becomes 9 greater when you flip it ?

Right there in the middle of the list . . . 34 ==> 43 .


 

Answer 2

$10x+y-the\ number\\\\ \left\{\begin{array}{ccc}x+y=7\\10y+x=10x+y+9\end{array}\right$