Question 87967
Let the cost per day for the room be $x and that of the car be $y.

As per the first plan, 3x+2y = 360

As per the second plan, 4x+3y = 500

So the system of equations are 

3x+2y = 360 -------------(1)

4x+3y = 500 -------------(2)

This is going to be solved by elimination.

(1) times 3 gives, 9x+6y = 1080

(2) times 2 gives, 8x+6y = 1000

Subtracting the above 2 equations, x = 80

Plugging in this value in (1), 
3(80)+2y = 360

240+2y = 360

==> 2y = 360 - 240

==> 2y = 120

==> y = 60

Thus the cost of a room per day is $80 and that for a car is $60