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. J
eremy 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.
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
