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?
I know that the answer is 34, and I know that one of the equations is x+y=7, but I do not know the other part to the system of equations. Thanks for helping me!!!

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 valueof 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 more10y + x = 10x + y + 9Just to make it neater and easier to handle, let's subtract 1x and 1y from each side:9y = 9x + 9Now.  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 + 99(7 - x) = 9x + 963 - 9x = 9x + 9Subtract 9 from each side:  54 - 9x = 9xAdd 9x to each side:    54 = 18xDivide each side by 18:    3 = xGreat.    y = 7 - x      y = 4The original number is  34 .    3 + 4 = 7Flip 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 . . . . . 6125 . . . . . 5234 . . . . . 4343 . . . . . 3452 . . . . . 25Do you see a number that becomes 9 greater when you flip it ?Right there in the middle of the list . . . 34 ==> 43 . 

[tex]10x+y-the\ number\\\\ \left\{\begin{array}{ccc}x+y=7\\10y+x=10x+y+9\end{array}\right[/tex]