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Felipe will rent a car for the weekend. He can choose one of two plans.The first plan has an initial fee of $55 and costs an additional $0.10 per mile down. The

Does not make since how they will ever reach the same amount because the .60 is clearly always going to be more no matter how many miles they are at. 

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Hey Angela, the simplest way to do this problem is setting both sides equal to each other.  You'll have 55 + .1x = .6x.  You can do this with virtually any "when will plan A be equal to plan B" type of problem.  Now, as we have our equation, let's solve for x; 55 = .6x - .1x => 55 = .5x.  Next, we divide 55/.5 = 110 = x.  Therefore, at 110 miles, the plans will be the same.

 

Another example of this would be say plan A is $45 down and $0.10 per mile, plan B is $65 down and $.05 a mile.  Which plan would be better if you were driving 50 miles.

Setting this up in the same way, we get 45 + .10x = 65 + .05x.  Combine like terms, and we get .05x = 20.  Dividing, we'll get x = 400 miles is where they're equal, with the Plan A better for short periods and plan B better for longer periods (lower slope).

Hey Angela -- if Felipe is going to drive a lot of miles, pay the fixed fee with the lower-mileage rate. Driving a few miles is better with the "no-fee" plan and higher-mileage rate. There is a certain number of miles where the cost is the same: 1st plan at 110 miles costs $55 + $11 or $66; 2nd plan at 60 cents/mile costs $66, too.   

Angela - I think you need to re-enter your question with a shorter title. This entry has cut off; all I can see of the question is "Felipe will rent a car for the weekend. He can choose one of two plans.The first plan has an initial fee of $55 and costs an additional $0.10 per mile down. The"