A bus can only have x passenagers seated and 12 passenagers standing. How many passenagers can y buses have? Leaving your answer in an algebraic expression.
On this bus, there is standing room for 12 passengers and an additional $x$ seats to accompany the $x$ passengers seating. This tells us that one buss will have room for $x + 12$ passengers.
Now, two busses will be able to carry twice this amount, which is $2(x + 12)$.
If we generalize this to any number $y$, then $y$ busses will be able to carry $y$ times this amount, which is $y(x + 12)$.
You can write your algebraic expression as $y(x + 12)$, or $yx + 12y$ if you distribute the $y$.
$P(y) = (x+12) * y$
One bus can hold a total of $x+12$ passengers, as $x$ passengers may be seated and $12$ can stand. Hence, $y$ buses can hold a total of $y$ times that amount.