Lesson 3.6: Solve Applications with Linear Inequalities

New Concept

Learn to translate real-world situations with limits, like budgets or minimums, into linear inequalities. By setting up and solving these inequalities, you can find the range of possible solutions for practical, everyday problems.

What’s next

Let's put this into practice. You'll work through interactive examples breaking down word problems and then solve a series of practice cards.

Property

The method we will use to solve applications with linear inequalities is very much like the one we used when we solved applications with equations.
We will read the problem and make sure all the words are understood. Next, we will identify what we are looking for and assign a variable to represent it.
We will restate the problem in one sentence to make it easy to translate into an inequality.
Then, we will solve the inequality.

Examples

• To rent a car, the company charges 90 dollars a week plus 0.20 dollars a mile. How many miles can you travel and stay within a 250 dollars budget? Let $m$ be miles. The inequality is $90 + 0.20m \leq 250$, which solves to $m \leq 800$. You can travel up to 800 miles.

• Emma's monthly income is 6,000 dollars. To qualify for an apartment, her income must be at least four times the rent. What is the highest rent she qualifies for? Let $r$ be the rent. The inequality is $6000 \geq 4r$, so $1500 \geq r$. The maximum rent is 1,500 dollars.

Property

Sometimes an application requires the solution to be a whole number, but the algebraic solution to the inequality is not a whole number. In that case, we must round the number up or down to find the correct answer that makes sense in the context of the problem.

Examples

• A teacher has a 5,000 dollars grant to buy classroom laptops that cost 315.50 dollars each. What is the maximum number of laptops she can buy? Let $n$ be the number of laptops. $315.50n \leq 5000$ gives $n \leq 15.84$. She can buy a maximum of 15 laptops.

• An elevator has a weight limit of 2,000 pounds. If the average person weighs 160 pounds, what is the maximum number of people allowed? Let $p$ be the number of people. $160p \leq 2000$ gives $p \leq 12.5$. The maximum is 12 people.

Property

Profit is the money that remains when the expenses have been subtracted from the money earned. To find the number of jobs or sales needed to make a certain amount of profit, use the inequality:
Revenue - Expenses $\geq$ Target Profit.

Examples

• A landscaper has monthly expenses of 1,500 dollars. If he charges 75 dollars per job, how many jobs must he do to earn a profit of at least 3,000 dollars? Let $j$ be jobs. $75j - 1500 \geq 3000$ solves to $j \geq 60$. He must do at least 60 jobs.

• A jewelry maker sells necklaces for 50 dollars each. Her monthly expenses are 800 dollars. How many necklaces must she sell to make a profit of at least 1,000 dollars? Let $n$ be necklaces. $50n - 800 \geq 1000$ solves to $n \geq 36$. She must sell at least 36 necklaces.

Property

There are many situations in which several quantities contribute to the total expense.
We must make sure to account for all the individual expenses when we solve problems like this.
The expenses must be less than or equal to the income.

Examples

• A student is planning a 4-day trip. Airfare is 400 dollars, food is 50 dollars per day, and the hotel is 80 dollars per night for 3 nights. He has 500 dollars saved. How many hours must he work at 20 dollars per hour to afford the trip? Let $h$ be hours. $400 + 50(4) + 80(3) \leq 500 + 20h$. This simplifies to $840 \leq 500 + 20h$, so $h \geq 17$. He must work at least 17 hours.

• A family is buying a computer. The monitor costs 200 dollars, the tower costs 450 dollars, and the warranty is 75 dollars. They have a 500 dollars gift card. If they save 50 dollars per week, how many weeks will it take to afford the computer? Let $w$ be weeks. $200 + 450 + 75 \leq 500 + 50w$. This simplifies to $725 \leq 500 + 50w$, so $w \geq 4.5$. They need to save for at least 5 weeks.