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Applied Linear Inequalities  Problem 1
So we are dealing with linear inequalities and word problems. Here is a word problem that says; you need to rent a car for a week. RentaWreck charges $28 a day with no mileage fee. Wheels by the week charges $108 a week plus 14 cents a mile. How many miles must you drive so RentaWreck is a better deal?
Well we can solve this using inequalities. So let’s start by setting up some systems of inequalities. Starting with RentaWreck, so RentaWreck charge is a flat fee of $28 a day with no mileage. So that’s simply $28 per day. And since we are renting for a full week, we simply multiply 7 times 28, which gives you $196.
Wheels by the week is a little trickier, let’s take a look. Wheels charge is $108 per week as a flat rate but it also charges 14 cents a mile. So if you drive one mile, that’s an additional 14 cents, 2 miles and you get charged 28 cents. So that’s a variable that we can write out like this 0.14x, x being the number of miles that you drive.
So now what we are going to want to do is determine where this price is larger than this price, because that’s what the question is asking us to do. So 196 is less than 108 plus 0.14x. Remember we are using less than because we want to determine when this number is larger than 196. And we are not using equal to because if it were equal, then it would be the same price and we wouldn’t get a better deal.
So we can solve this algebraically. Simply subtract 108 from both sides, and we are left with 88 is less than 0.14x. Divide both sides by 0.14 and we end up with 628.57 miles is less than x. So what this is telling us is that we need to plan ahead.
If we think we are going to drive more than 628 miles then RentaWreck is the best deal. But if we are going to drive less than 628.57 miles, then wheels by the week is the best deal. And this is solving a word problem using linear inequalities.
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Applied Linear Inequalities
Problem 1 4,463 viewsYou need to rent a car for a week. RentaWreck charges $28/day with no mileage fee. Wheelsbytheweek charges $108/week plus 14 cents a mile.
How many miles must you drive so RentaWreck is a better deal? 
Applied Linear Inequalities
Problem 2 619 views 
Applied Linear Inequalities
Problem 3 696 views
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