The sum of the digits of a two-digit number is 11. the number obtained by interchanging the digits exceeds twice the original number by 34. what is the original number?

Cool problem

x and y are the digits
x+y=11
yx=2(xy)+34
I'm not sure how to solve this using legit math so use trial and error

the 1 digit numbers that could be possible are
9,2
8,3
7,4
6,5
5,6
4,7
3,8
2,9
all of them add up to 11

the pairs obtained by swiching the number are
92 and 29
83 and 38
74 and 47
65 and 56

we find which ones are equal when we subtract 34 from the larger number and divide by 2
92-34=58
58/2=29
92 and 29 work

83-34=49
49/2=24.5 not equal to 38 not work

74-34=40
40/2=20 47 not equal to 20 so doesn't work

65-34=31
31/2=15.5 not equal to 56 so doesn't work

the origonal number is 29

F(2) = 0f(-2) = 0due to the way the equation is, answers with 0 are correct.

What is your question again i don't understand