The two numbers are 41 and 8. You can set up the problem as two equations. First, x = 33 + y for "one number (x) is (=) 33 more (33 +) than another number (y)." You can set up the other equation as (x-1)/5y or (x-1) = 5y. You can substitute 33 + y for every x in the second equation and solve for y. You would get y = 8, then you can use the y value in the first equation to solve for x to get x = 41.
