The task is to find a double digit number which is 19 times greater than it's second digit

I made an equation which looked like this: 10a+b=19b (a is the first digit and b is the second one)

but there are two unknown values(a,b) so i need another equation but how would it look like because i cant think of any way i can make it.

The answer is easy to figure out because its just a double digit but how would i prove that i am not guessing. The answer is 95.

Another equation would be helpful, but it is not necessary. Let's see what else you know about these 'a' and 'b'?

10a + b = 19b -- That's where you started.

We're in a Base 10 Positional System, so both 'a' and 'b' must be from {0,1,2,3,4,...9}

Neither can be zero (0), since a = 0 makes a 1-digit number and b = 0 doesn't leave much for 19*b. Wait! Is 10 * 0 + 0 = 19*0? Interesting idea that we can keep in the back of our head.

b CANNOT be in {6, 7, 8, 9}, since 19 multiplied by any of these would result in a 3-digit number for 10a + b.

That just leaves us with b in {1, 2, 3, 4, 5}

Also, when multiplying 'b' by 9 (the one's digit of 19), the digit must remain the same.

1*9 = 9 -- Nope -- 1 is not 9

2*9 = 18 -- Nope -- 2 is not 8

3*9 = 27 -- Nope -- 3 is not 7

4*9 = 36 -- Nope -- 4 is not 6

I see a pattern!

5*9 = 45 -- We could have something!

Thinking. Hunting. Eliminating. Thinking. Reason it out.