Jeremy is taking a photography class, but he doesn't own a camera. the class organizers will rent him a camera for $6 per day. jeremy can spend up to $50 for the class, but he has to pay $15 to register for the class as well as rent the camera. the number of days, d, that jeremy can rent the camera is represented by the inequality 6d + 15 < 50. select the number of days jeremy can rent the camera with the money he has.

Solving the given inequality for d, we get...

6d + 15 < 50
6d + 15-15 < 50-15 <<-- subtract 15 from both sides
6d < 35
6d/6 < 35/6 <<--- divide both sides by 6
d < 5.83

Which means that d can be any of the values in this set: {0, 1, 2, 3, 4, 5}

The smallest d can be is 0. In this scenario, Jeremy pays the $15 registration but doesn't rent the camera at all
The largest d can be is 5. In this scenario, Jeremy rents the camera for 5 days
Any larger value of d is not allowed as it would make the total cost go over $50
Notice how I'm rounding down regardless how close 5.83 is to 6