Take it easy!

Let the one's digit be $Y$ and ten's digit be $X$ .
●According to first statement.
$X>Y$ .
●According to second statement.
$X+Y=2(X-Y)$
$X-3Y=0$
$X=3Y$ .
This statement gives that ten's digit should be 3 times the one's digit.
Now, putting $Y$ as 1,2,3 but if you put $Y$ greater than or equal to 4 the number will be of three digit or more.
$X=3×1=3$
$X=3×2=6$
$X=3×3=9$ .
Now possible two digit numbers which satisfies both conditions are $31,62,93$ .
Now, their sum becomes
$31+62+93=\boxed{186}$ .