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.
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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 .