My solution:
Considering two-digit numbers, their difference can be written as $(10a + b) - (10b + a)$, where $a$ and $b$ are positive integers representing the first and second digits of the larger number, respectively. Rewriting the expression gives $9(a-b)$. As their difference must end with 2, we can write it as $10k + 2$, where $k$ is another positive integer. The whole equation, then, is
$9(a-b) = 10k + 2$
Substituting $n$ for $a-b$ yields $9n = 10k + 2$, meaning we're looking for a positive multiple of 9 ending with 2. The smallest such is 72, and the rest take the form $72+90m$, where $m$ is yet another positive integer. Rewriting as an equation:
$9n = 72 + 90m$
$9n = 8*9 + 90m$
$9n = 9(8+10m)$
$n = 8+10m$
$a-b = 8+10m$
$(10a+b)+(10b+a) = 8+10m + 12b + 10a$
If by "lowest possible" the author meant their sum and you're not in the same tent as your twin, we can consider $m$ and $b$ to be 0*, giving:
$(10a+0)+(10*0+a) = 8+10*0 + 12*0 + 10a$
$10a + a = 8 + 10a$
$a=8$
The numbers, then, are 80 and 08
*Solving for b instead gives $b=-8$