problem solving involving linear inequalities

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3.6: Solve Applications with Linear Inequalities

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Learning Objectives

By the end of this section, you will be able to:

• Solve applications with linear inequalities

Before you get started, take this readiness quiz.

• Write as an inequality: x is at least 30. If you missed this problem, review Exercise 2.7.34 .
• Solve \(8−3y<41\). If you missed this problem, review Exercise 2.7.22 .

Solve Applications with Linear Inequalities

Many real-life situations require us to solve inequalities. In fact, inequality applications are so common that we often do not even realize we are doing algebra. For example, how many gallons of gas can be put in the car for $20? Is the rent on an apartment affordable? Is there enough time before class to go get lunch, eat it, and return? How much money should each family member’s holiday gift cost without going over budget?

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.

Example \(\PageIndex{1}\)

Emma got a new job and will have to move. Her monthly income will be $5,265. To qualify to rent an apartment, Emma’s monthly income must be at least three times as much as the rent. What is the highest rent Emma will qualify for?

\(\begin{array} {ll} {\textbf{Step 1. Read} \text{ the problem.}} &{} \\ {\textbf{Step 2. Identify} \text{ what we are looking for.}} &{\text{the highest rent Emma will qualify for}} \\ {\textbf{Step 3. Name} \text{ what we are looking for.}} &{} \\ {} &{\text{Let r = rent}} \\ {\text{Choose a variable to represent that quantity.}} &{} \\{\textbf{Step 4. Translate} \text{ into an inequality.}} &{} \\{} &{\text{Emma’s monthly income must be at least}} \\ {\text{First write a sentence that gives the information}} &{\text{three times the rent.}} \\ {\text{to find it.}} &{} \\\\ {\textbf{Step 5. Solve} \text{ the inequality.}} &{5265 \geq 3r} \\ {\text{Remember, } a > x\text{ has the same meaning}} &{1755 \geq r} \\ {\text{as }x < a} &{r \leq 1755} \\ {\textbf{Step 6. Check} \text{ the answer in the problem}} &{} \\ {\text{and make sure it makes sense.}} &{} \\ {\text{A maximum rent of \$1,755 seems}} &{} \\ {\text{reasonable for an income of \$5,265.}} &{} \\ {\textbf{Step 7. Answer} \text{ the answer in the problem}} &{\text{the question with a}} \\ {\text{complete sentence.}} &{\text{The maximum rent is \$1,755.}} \end{array}\)

Try It \(\PageIndex{2}\)

Alan is loading a pallet with boxes that each weighs 45 pounds. The pallet can safely support no more than 900 pounds. How many boxes can he safely load onto the pallet?

There can be no more than 20 boxes.

Try It \(\PageIndex{3}\)

The elevator in Yehire’s apartment building has a sign that says the maximum weight is 2,100 pounds. If the average weight of one person is 150 pounds, how many people can safely ride the elevator?

A maximum of 14 people can safely ride in the elevator.

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 algebraic solution to a whole number. The context of the application will determine whether we round up or down. To check applications like this, we will round our answer to a number that is easy to compute with and make sure that number makes the inequality true.

Example \(\PageIndex{4}\)

Dawn won a mini-grant of $4,000 to buy tablet computers for her classroom. The tablets she would like to buy cost $254.12 each, including tax and delivery. What is the maximum number of tablets Dawn can buy?

\(\begin{array} {ll} {\textbf{Step 1. Read} \text{ the problem.}} &{} \\ {\textbf{Step 2. Identify} \text{ what we are looking for.}} &{\text{the maximum number of tablets Dawn can buy}} \\ {\textbf{Step 3. Name} \text{ what we are looking for.}} &{} \\ {} &{\text{Let n = the number of tablets.}} \\ {\text{Choose a variable to represent that quantity.}} &{} \\{\textbf{Step 4. Translate.} \text{ write a sentence that}} &{} \\{\text{gives the information to find it.}} &{$254.12\text{ times the number of tablets is no}} \\ {} &{\text{more than \$4000.}} \\ {\text{Translate into an inequality.}} &{254.12n \leq 4000} \\ {\textbf{Step 5. Solve} \text{ the inequality.}} &{n \leq 15.74} \\ {\text{But n must be a whole number of tablets,}} &{} \\ {\text{so round to 15.}} &{n \leq 15}\\ \\{\textbf{Step 6. Check} \text{ the answer in the problem}} &{} \\ {\text{and make sure it makes sense.}} &{} \\ {\text{Rounding down the price to \$250,}} &{} \\ {\text{15 tablets would cost \$3750, while}} &{} \\ {\text{16 tablets would be \$4000. So a}} &{} \\{\text{maximum of 15 tablets at \$254.12}} &{} \\ {\text{seems reasonable.}} &{} \\{\textbf{Step 7. Answer} \text{ the answer in the problem}} &{\text{the question with a}} \\ {\text{complete sentence.}} &{\text{Dawn can buy a maximum of 15 tablets.}} \end{array}\)

Try It \(\PageIndex{5}\)

Angie has $20 to spend on juice boxes for her son’s preschool picnic. Each pack of juice boxes costs $2.63. What is the maximum number of packs she can buy?

seven packs

Try It \(\PageIndex{6}\)

Daniel wants to surprise his girlfriend with a birthday party at her favorite restaurant. It will cost $42.75 per person for dinner, including tip and tax. His budget for the party is $500. What is the maximum number of people Daniel can have at the party?

Example \(\PageIndex{7}\)

Pete works at a computer store. His weekly pay will be either a fixed amount, $925, or $500 plus 12% of his total sales. How much should his total sales be for his variable pay option to exceed the fixed amount of $925?

\(\begin{array} {ll} {\textbf{Step 1. Read} \text{ the problem.}} &{} \\ {\textbf{Step 2. Identify} \text{ what we are looking for.}} &{\text{the total sales needed for his variable pay}} \\ {} &{\text{option to exceed the fixed amount of \$925}} \\ {\textbf{Step 3. Name} \text{ what we are looking for.}} &{} \\ {} &{\text{Let s = the total sales.}} \\ {\text{Choose a variable to represent that quantity.}} &{} \\{\textbf{Step 4. Translate.} \text{ write a sentence that}} &{} \\{\text{gives the information to find it.}} &{$500\text{ plus 12% of total sales is more than \$925.}} \\ {\text{Translate into an inequality. Remember to}} &{500 + 0.12s > 925} \\{\text{convert the percent to a decimal.}} &{} \\\\ {\textbf{Step 5. Solve} \text{ the inequality.}} &{0.12s > 425} \\ {} &{s > 3541.\overline{66}} \\ \\ \\{\textbf{Step 6. Check} \text{ the answer in the problem}} &{} \\ {\text{and make sure it makes sense.}} &{} \\ {\text{Rounding down the price to \$250,}} &{} \\ {\text{15 tablets would cost \$3750, while}} &{} \\ {\text{If we round the total sales up to}} &{} \\{\text{\$4000, we see that}} &{} \\ {\text{500+0.12(4000) = 980, which is more}} &{} \\ {\text{than \$925.}} &{} \\{\textbf{Step 7. Answer} \text{ the the question with a complete sentence.}} &{\text{The total sales must be more than \$3541.67}} \end{array}\)

Try It \(\PageIndex{8}\)

Tiffany just graduated from college and her new job will pay her $20000 per year plus 2% of all sales. She wants to earn at least $100000 per year. For what total sales will she be able to achieve her goal?

at least $4000000

Try It \(\PageIndex{9}\)

Christian has been offered a new job that pays $24000 a year plus 3% of sales. For what total sales would this new job pay more than his current job which pays $60000?

at least $1200000

Example \(\PageIndex{10}\)

Sergio and Lizeth have a very tight vacation budget. They plan to rent a car from a company that charges $75 a week plus $0.25 a mile. How many miles can they travel and still keep within their $200 budget?

\(\begin{array} {ll} {\textbf{Step 1. Read} \text{ the problem.}} &{} \\ {\textbf{Step 2. Identify} \text{ what we are looking for.}} &{\text{the number of miles Sergio and Lizeth can travel}} \\ {\textbf{Step 3. Name} \text{ what we are looking for.}} &{} \\ {} &{\text{Let m = the number of miles.}} \\ {\text{Choose a variable to represent that quantity.}} &{} \\{\textbf{Step 4. Translate.} \text{ write a sentence that}} &{\text{\$75 plus 0.25 times the number of miles is}} \\{\text{gives the information to find it.}} &{\text{ less than or equal to \$200.}} \\ {\text{Translate into an inequality. }} &{75 + 25m \leq 200} \\\\ {\textbf{Step 5. Solve} \text{ the inequality.}} &{0.25m \leq 125} \\ {} &{m \leq 500 \text{ miles}} \\ \\ \\{\textbf{Step 6. Check} \text{ the answer in the problem}} &{} \\ {\text{and make sure it makes sense.}} &{} \\ {\text{Yes, 75 + 0.25(500) = 200.}} & {}\\{\textbf{Step 7. Answer} \text{ the the question with a complete sentence.}} &{\text{Sergio and Lizeth can travel 500 miles}} \\ {} &{\text{and still stay on budget.}} \end{array}\)

Try It \(\PageIndex{11}\)

Taleisha’s phone plan costs her $28.80 a month plus $0.20 per text message. How many text messages can she use and keep her monthly phone bill no more than $50?

no more than 106 text messages

Try It \(\PageIndex{12}\)

Rameen’s heating bill is $5.42 per month plus $1.08 per therm. How many therms can Rameen use if he wants his heating bill to be a maximum of $87.50?

no more than 76 therms

A common goal of most businesses is to make a profit. Profit is the money that remains when the expenses have been subtracted from the money earned. In the next example, we will find the number of jobs a small businessman needs to do every month in order to make a certain amount of profit .

Example \(\PageIndex{13}\)

Elliot has a landscape maintenance business. His monthly expenses are $1,100. If he charges $60 per job, how many jobs must he do to earn a profit of at least $4,000 a month?

\(\begin{array} {ll} {\textbf{Step 1. Read} \text{ the problem.}} &{} \\ {\textbf{Step 2. Identify} \text{ what we are looking for.}} &{\text{the number of jobs Elliot needs}} \\ {\textbf{Step 3. Name} \text{ what we are looking for.}} &{} \\ {\text{Choose a variable to represent it}} &{\text{Let j = the number of jobs.}} \\{\textbf{Step 4. Translate.} \text{ write a sentence that}} &{\text{\$60 times the number of jobs minus \$1,100 is at least \$4,000.}} \\{\text{gives the information to find it.}} &{\text{ less than or equal to \$200.}} \\ {\text{Translate into an inequality. }} &{60j - 1100 \geq 4000} \\\\ {\textbf{Step 5. Solve} \text{ the inequality.}} &{60j \geq 5100} \\ {} &{j \geq 85\text{ jobs}} \\ \\{\textbf{Step 6. Check} \text{ the answer in the problem}} &{} \\ {\text{and make sure it makes sense.}} &{} \\ {\text{If Elliot did 90 jobs, his profit would be}} & {}\\ {\text{60(90)−1,100,or \$4,300. This is}} &{} \\ {\text{more than \$4,000.}} &{} \\{\textbf{Step 7. Answer} \text{ the the question with a complete sentence.}} &{\text{Elliot must work at least 85 jobs.}} \end{array}\)

Try It \(\PageIndex{14}\)

Caleb has a pet sitting business. He charges $32 per hour. His monthly expenses are $2272. How many hours must he work in order to earn a profit of at least $800 per month?

at least 96 hours

Try It \(\PageIndex{15}\)

Felicity has a calligraphy business. She charges $2.50 per wedding invitation. Her monthly expenses are $650. How many invitations must she write to earn a profit of at least $2800 per month?

at least 1380 invitations

Sometimes life gets complicated! 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.

Example \(\PageIndex{16}\)

Brenda’s best friend is having a destination wedding and the event will last 3 days. Brenda has $500 in savings and can earn $15 an hour babysitting. She expects to pay $350 airfare, $375 for food and entertainment and $60 a night for her share of a hotel room. How many hours must she babysit to have enough money to pay for the trip?

\(\begin{array} {ll} {\textbf{Step 1. Read} \text{ the problem.}} &{} \\ {\textbf{Step 2. Identify} \text{ what we are looking for.}} &{\text{the number of hours Brenda must babysit}} \\ {\textbf{Step 3. Name} \text{ what we are looking for.}} &{} \\ {\text{Choose a variable to represent that quantity.}} &{\text{Let h = the number of hours.}} \\{\textbf{Step 4. Translate.} \text{ write a sentence that}} &{} \\{\text{gives the information to find it.}} &{} \\ {} &{\text{The expenses must be less than or equal to}} \\ {} &{\text{the income. The cost of airfare plus the}} \\ {} &{\text{cost of food and entertainment and the}} \\ {} &{\text{hotel bill must be less than or equal to the savings}} \\ {} &{\text{plus the amount earned babysitting.}} \\ {\text{Translate into an inequality. }} &{\$350 + \$375 + \$60(3) \leq \$500 + \$15h} \\\\ {\textbf{Step 5. Solve} \text{ the inequality.}} &{905 \leq 500 + 15h} \\{} &{405 \leq 15h} \\ {} &{27 \leq h} \\ {} &{h \geq 27} \\ \\{\textbf{Step 6. Check} \text{ the answer in the problem}} &{} \\ {\text{and make sure it makes sense.}} &{} \\ {\text{We substitute 27 into the inequality.}} & {}\\{905 \leq 500 + 15h} &{} \\ {905 \leq 500 + 15(27)} &{} \\ {905 \leq 905} &{} \\ \\{\textbf{Step 7. Answer} \text{ the the question with a complete sentence.}} &{\text{Brenda must babysit at least 27 hours.}} \end{array}\)

Try It \(\PageIndex{17}\)

Malik is planning a 6-day summer vacation trip. He has $840 in savings, and he earns $45 per hour for tutoring. The trip will cost him $525 for airfare, $780 for food and sightseeing, and $95 per night for the hotel. How many hours must he tutor to have enough money to pay for the trip?

at least 23 hours

Try It \(\PageIndex{18}\)

Josue wants to go on a 10-day road trip next spring. It will cost him $180 for gas, $450 for food, and $49 per night for a motel. He has $520 in savings and can earn $30 per driveway shoveling snow. How many driveways must he shovel to have enough money to pay for the trip?

at least 20 driveways

Key Concepts

• Read the problem.
• Identify what we are looking for.
• Name what we are looking for. Choose a variable to represent that quantity.
• Translate. Write a sentence that gives the information to find it. Translate into an inequality.
• Solve the inequality.
• Check the answer in the problem and make sure it makes sense.
• Answer the question with a complete sentence.

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Examples of Solving Harder Linear Inequalities

Intro & Formatting Worked Examples Harder Examples & Word Prob's

Once you'd learned how to solve one-variable linear equations, you were then given word problems. To solve these problems, you'd have to figure out a linear equation that modelled the situation, and then you'd have to solve that equation to find the answer to the word problem.

Content Continues Below

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Solving Inequalities

So, now that you know how to solve linear inequalities — you guessed it! — they give you word problems.

• The velocity of an object fired directly upward is given by V = 80 − 32 t , where the time t is measured in seconds. When will the velocity be between 32 and 64 feet per second (inclusive)?

This question is asking when the velocity, V , will be between two given values. So I'll take the expression for the velocity,, and put it between the two values they've given me. They've specified that the interval of velocities is inclusive, which means that the interval endpoints are included. Mathematically, this means that the inequality for this model will be an "or equal to" inequality. Because the solution is a bracket (that is, the solution is within an interval), I'll need to set up a three-part (that is, a compound) inequality.

I will set up the compound inequality, and then solve for the range of times t :

32 ≤ 80 − 32 t ≤ 64

32 − 80 ≤ 80 − 80 − 32 t ≤ 64 − 80

−48 ≤ −32 t ≤ −16

−48 / −32 ≥ −32 t / −32 ≥ −16 / −32

1.5 ≥ t ≥ 0.5

Note that, since I had to divide through by a negative, I had to flip the inequality signs.

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Note also that you might (as I do) find the above answer to be more easily understood if it's written the other way around, with "less than" inequalities.

And, because this is a (sort of) real world problem, my working should show the fractions, but my answer should probably be converted to decimal form, because it's more natural to say "one and a half seconds" than it is to say "three-halves seconds". So I convert the last line above to the following:

0.5 ≤ t ≤ 1.5

Looking back at the original question, it did not ask for the value of the variable " t ", but asked for the times when the velocity was between certain values. So the actual answer is:

The velocity will be between 32 and 64 feet per second between 0.5 seconds after launch and 1.5 seconds after launch.

Okay; my answer above was *extremely* verbose and "complete"; you don't likely need to be so extreme. You can probably safely get away with saying something simpler like, "between 0.5 seconds and 1.5 seconds". Just make sure that you do indeed include the approprioate units (in this case, "seconds").

Always remember when doing word problems, that, once you've found the value for the variable, you need to go back and re-read the problem to make sure that you're answering the actual question. The inequality 0.5 ≤  t  ≤ 1.5 did not answer the actual question regarding time. I had to interpret the inequality and express the values in terms of the original question.

• Solve 5 x + 7 < 3( x  + 1) .

First I'll multiply through on the right-hand side, and then solve as usual:

5 x + 7 < 3( x + 1)

5 x + 7 < 3 x + 3

2 x + 7 < 3

2 x < −4

x < −2

In solving this inequality, I divided through by a positive 2 to get the final answer; as a result (that is, because I did *not* divide through by a minus), I didn't have to flip the inequality sign.

• You want to invest $30,000 . Part of this will be invested in a stable 5% -simple-interest account. The remainder will be "invested" in your father's business, and he says that he'll pay you back with 7% interest. Your father knows that you're making these investments in order to pay your child's college tuition with the interest income. What is the least you can "invest" with your father, and still (assuming he really pays you back) get at least $1900 in interest?

First, I have to set up equations for this. The interest formula for simple interest is I = Prt , where I is the interest, P is the beginning principal, r is the interest rate expressed as a decimal, and t is the time in years.

Since no time-frame is specified for this problem, I'll assume that t  = 1 ; that is, I'll assume (hope) that he's promising to pay me at the end of one year. I'll let x be the amount that I'm going to "invest" with my father. Then the rest of my money, being however much is left after whatever I give to him, will be represented by "the total, less what I've already given him", so 30000 −  x will be left to invest in the safe account.

Then the interest on the business investment, assuming that I get paid back, will be:

I = ( x )(0.07)(1) = 0.07 x

The interest on the safe investment will be:

(30 000 − x )(0.05)(1) = 1500 − 0.05 x

The total interest is the sum of what is earned on each of the two separate investments, so my expression for the total interest is:

0.07 x + (1500 − 0.05 x ) = 0.02 x + 1500

I need to get at least $1900 ; that is, the sum of the two investments' interest must be greater than, or at least equal to, $1,900 . This allows me to create my inequality:

0.02 x + 1500 ≥ 1900

0.02 x ≥ 400

x ≥ 20 000

That is, I will need to "invest" at least $20,000 with my father in order to get $1,900 in interest income. Since I want to give him as little money as possible, I will give him the minimum amount:

I will invest $20,000 at 7% .

• An alloy needs to contain between 46% copper and 50% copper. Find the least and greatest amounts of a 60% copper alloy that should be mixed with a 40% copper alloy in order to end up with thirty pounds of an alloy containing an allowable percentage of copper.

This is similar to a mixture word problem , except that this will involve inequality symbols rather than "equals" signs. I'll set it up the same way, though, starting with picking a variable for the unknown that I'm seeking. I will use x to stand for the pounds of 60% copper alloy that I need to use. Then 30 −  x will be the number of pounds, out of total of thirty pounds needed, that will come from the 40% alloy.

Of course, I'll remember to convert the percentages to decimal form for doing the algebra.

How did I get those values in the bottom right-hand box? I multiplied the total number of pounds in the mixture ( 30 ) by the minimum and maximum percentages ( 46% and 50% , respectively). That is, I multiplied across the bottom row, just as I did in the " 60% alloy" row and the " 40% alloy" row, to get the right-hand column's value.

The total amount of copper in the mixture will be the sum of the copper contributed by each of the two alloys that are being put into the mixture. So I'll add the expressions for the amount of copper from each of the alloys, and place the expression for the total amount of copper in the mixture as being between the minimum and the maximum allowable amounts of copper:

13.8 ≤ 0.6 x + (12 − 0.4 x ) ≤ 15

13.8 ≤ 0.2 x + 12 ≤ 15

1.8 ≤ 0.2 x ≤ 3

9 ≤ x ≤ 15

Checking back to my set-up, I see that I chose my variable to stand for the number of pounds that I need to use of the 60% copper alloy. And they'd only asked me for this amount, so I can ignore the other alloy in my answer.

I will need to use between 9 and 15 pounds of the 60% alloy.

Per yoozh, I'm verbose in my answer. You can answer simply as " between 9 and 15 pounds ".

• Solve 3( x − 2) + 4 ≥ 2(2 x − 3)

First I'll multiply through and simplify; then I'll solve:

3( x − 2) + 4 ≥ 2(2 x − 3)

3 x − 6 + 4 ≥ 4 x − 6

3 x − 2 ≥ 4 x − 6

−2 ≥ x − 6            (*)

Why did I move the 3 x over to the right-hand side (to get to the line marked with a star), instead of moving the 4 x to the left-hand side? Because by moving the smaller term, I was able to avoid having a negative coefficient on the variable, and therefore I was able to avoid having to remember to flip the inequality when I divided through by that negative coefficient. I find it simpler to work this way; I make fewer errors. But it's just a matter of taste; you do what works for you.

Why did I switch the inequality in the last line and put the variable on the left? Because I'm more comfortable with inequalities when the answers are formatted this way. Again, it's only a matter of taste. The form of the answer in the previous line, 4 ≥ x , is perfectly acceptable.

As long as you remember to flip the inequality sign when you multiply or divide through by a negative, you shouldn't have any trouble with solving linear inequalities.

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Grade 8 Mathematics Module: “Solving Problems Involving Linear Inequalities in Two Variables”

This Self-Learning Module (SLM) is prepared so that you, our dear learners, can continue your studies and learn while at home. Activities, questions, directions, exercises, and discussions are carefully stated for you to understand each lesson.

Each SLM is composed of different parts. Each part shall guide you step-by-step as you discover and understand the lesson prepared for you.

Pre-tests are provided to measure your prior knowledge on lessons in each SLM. This will tell you if you need to proceed on completing this module or if you need to ask your facilitator or your teacher’s assistance for better understanding of the lesson. At the end of each module, you need to answer the post-test to self-check your learning. Answer keys are provided for each activity and test. We trust that you will be honest in using these.

Please use this module with care. Do not put unnecessary marks on any part of this SLM. Use a separate sheet of paper in answering the exercises and tests. And read the instructions carefully before performing each task.

If you have any questions in using this SLM or any difficulty in answering the tasks in this module, do not hesitate to consult your teacher or facilitator.

In this module, you will be acquainted with key concepts of solving problems involving linear inequalities in two variables. You are given the opportunity to use your prior knowledge and skills in translating mathematical expressions to verbal phrase and vice-versa, then solve problems involving real-life situations. The lesson is arranged accordingly to suit to your learning needs.

This module contains:

Lesson 1- Solving Problems Involving Linear Inequalities in Two Variables

After going through this module, you are expected to:

1. translate statements into mathematical expressions.

2. solve problems involving linear inequalities in two variables; and

3. apply linear inequalities in two variables in real-life situation.

Grade 8 Mathematics Quarter 2 Self-Learning Module: “Solving Problems Involving Linear Inequalities in Two Variables”

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• Inequalities
• 1. Properties of Inequalities

2. Solving Linear Inequalities

• 3. Solving Non-Linear Inequalties
• 4. Inequalities Involving Absolute Values

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Get help with your math queries:

The procedure for solving linear inequalities in one variable is similar to solving basic equations. (See Solving Equations .)

We need to be careful about the sense of the equality when multiplying or dividing by negative numbers.

Following are several examples of solving equations involving inequalities.

Solve x + 2

We need to subtract `2` from both sides of the inequality.

`x+2<4` `x<4-2` `x<2`

The graph of this solution is as follows:

Solve `x/2>4`

We need to multiply both sides of the inequality by `2`.

`x/2>4` `x>4xx2` `x>8`

Here's the graph of this solution:

Solve 2 x ≤ 4

We need to divide both sides of the inequality by `2`.

`2x<=4` `x<=4/2` `x<=2`

Solve the inequality 3 − 2 x ≥ 15

In this example, we need to subtract 3 from both sides; then divide both sides by `-2` (remembering to change the direction of the inequality).

`3-2x>=15` `-2x>=15-3` `-2x>=12` `x<=12/(-2)` `x<=-6`

(Note the change in sense due to dividing by a negative number)

Check: Always check your solution and you can be sure your answer is correct.

In this case, any number less than `-6` should "work" in the original equation, and any number bigger than `-6` should fail.

Let's take `x = -10` (a convenient number less than `-6`)

LHS `= 3 − 2(-10) = 3 + 20 = 23`. This is more than `15` so it is true.

Now let's take `x = 0` (a convenient number greater than `-6`)

LHS `= 3 − 2(0) = 3`. This is NOT more than `15`, which is what we hoped for.

So we can be sure our answer is correct.

Solve the inequality `3/2(1-x)>1/4-x`

`3/2(1-x)>1/4-x`

Multiplying both sides by 4 gives us:

`6(1-x)>1-4x` ` 6-6x>1-4x` `-6x+4x>1-6` `-2x> -5` `x<5/2`

(Note the change in sense in the last line, due to dividing by a negative number).

Check: Taking x = 0 (which should work):

`"LHS" = 3/2(1 − 0) = 3/2` `"RHS" = 1/4` It is TRUE that `3/2 > 1/4`, so that is good.

Now we take x = 3 (a convenient number bigger than 5/2, which should not work):

`"LHS" = 3/2(1 − 3) = -3` `"RHS" = 1/4 − 3 = -2 3/4` It is NOT true that `-3 > -2 3/4` and so `x=3` fails, as we hoped.

We can be sure our answer (`x<5/2`) is correct.

Inequalities with Three Members

Solve −1 < 2 x + 3 < 6

`- 1 < 2x + 3 < 6`

Subtract `3` from all 3 sides

`- 1 - 3 < 2x + 3 - 3 < 6 - 3` `- 4 < 2x < 3`

Divide all sides by `2`

`-2 < x < 3/2`

The solution graph:

Solve 2 x < x − 4 ≤ 3 x + 8

Another way of solving more complicated inequalities with 3 members (sides) is to rewrite the inequality as

`2x < x - 4` and `x - 4 ≤ 3x + 8`

Then solving each of these inequalities, we obtain:

LHS inequality:

`x < - 4`

RHS inequality:

`x - 4 ≤ 3x + 8` `- 4 ≤ 2x + 8` `- 12 ≤ 2x` `x ≥ -6 `

Bearing in mind that the final solution has to satisfy both inequalities, we obtain:

`x < -4` and `x ≥ -6`

These two portions appear as:

On considering the region where the two portions intersect, we get

`-6 ≤ x < -4`

This is the final solution graph:

Solve the following inequalities for x :

1. Solve 3 − 3 x < − 1

`3 - 3x < -1`

`x> 4/3 ~~ 1.333...`

Here is the solution graph:

2. Solve −2( x + 4) > 1 − 5 x

`-2(x + 4) > 1 - 5x`

`3x - 8 > 1`

`3x > 9`

3. Solve `x/5-2>2/3(x+3)`

`x/5-2>2/3(x+3)`

Multiply throughout by 5:

`x-10>10/3(x+3)`

Multiply throughout by 3:

`3x-30>10(x+3)` `3x-30>10x+30` `3x>10x+60` `-7x>60` `x<(-60)/7~~-8.57`

4. Solve x − 1 < 2 x + 2 < 3 x + 1

We need to find the intersection of the "true" values.

`x - 1 < 2x + 2` and `2x + 2 < 3x + 1`

`x < 2x + 3` and `2x < 3x - 1`

`x > -3` and `x > 1`

These 2 inequalities on the one axis appear as follows:

The intersection of these 2 regions is `x > 1`.

Always check your answers!

Here's the final solution graph:

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• School Guide
• Class 11 Syllabus
• Class 11 Revision Notes
• Maths Notes Class 11
• Physics Notes Class 11
• Chemistry Notes Class 11
• Biology Notes Class 11
• NCERT Solutions Class 11 Maths
• RD Sharma Solutions Class 11
• Math Formulas Class 11
• CBSE Class 11 Maths Notes

Chapter 1: Sets

• Representation of a Set
• Types Of Sets
• Universal Sets
• Venn Diagram
• Operations on Sets
• Union of Sets

Chapter 2: Relations & Functions

• Cartesian Product of Sets
• Relation and Function
• Introduction to Domain and Range
• Piecewise Function
• Range of a Function

Chapter 3: Trigonometric Functions

• Measuring Angles
• Trigonometric Functions
• Trigonometric Functions of Sum and Difference of Two Angles

Chapter 4: Principle of Mathematical Induction

• Principle of Mathematical Induction

Chapter 5: Complex Numbers and Quadratic Equations

• Complex Numbers
• Algebra of Real Functions
• Algebraic Operations on Complex Numbers | Class 11 Maths
• Polar Representation of Complex Numbers
• Absolute Value of a Complex Number
• Conjugate of Complex Numbers
• Imaginary Numbers

Chapter 6: Linear Inequalities

• Compound Inequalities
• Algebraic Solutions of Linear Inequalities in One Variable and their Graphical Representation - Linear Inequalities | Class 11 Maths
• Graphical Solution of Linear Inequalities in Two Variables

Solving Linear Inequalities Word Problems

Chapter 7: permutations and combinations.

• Fundamental Principle of Counting
• Permutation
• Combinations

Chapter 8: Binomial Theorem

• Binomial Theorem
• Pascal's Triangle

Chapter 9: Sequences and Series

• Sequences and Series
• General and Middle Terms - Binomial Theorem - Class 11 Maths
• Arithmetic Series
• Arithmetic Sequence
• Geometric Progression (GP) | Formula and Properties
• Geometric Series
• Arithmetic Progression and Geometric Progression
• Special Series - Sequences and Series | Class 11 Maths

Chapter 10: Straight Lines

• Slope of a Line
• Introduction to Two-Variable Linear Equations in Straight Lines
• Forms of Two-Variable Linear Equations - Straight Lines | Class 11 Maths
• Point-slope Form - Straight Lines | Class 11 Maths
• Slope Intercept Form
• Writing Slope-Intercept Equations
• Standard Form of a Straight Line
• X and Y Intercept
• Graphing slope-intercept equations - Straight Lines | Class 11 Maths

Chapter 11: Conic Sections

• Conic Sections
• Equation of a Circle
• Focus and Directrix of a Parabola
• Identifying Conic Sections from their Equation

Chapter 12: Introduction to Three-dimensional Geometry

• Coordinate Axes and Coordinate planes in 3D space
• 3D Distance Formula

Chapter 13: Limits and Derivatives

• Formal Definition of Limits
• Strategy in Finding Limits
• Determining Limits using Algebraic Manipulation
• Limits of Trigonometric Functions
• Properties of Limits
• Limits by Direct Substitution
• Estimating Limits from Graphs
• Estimating Limits from Tables
• Sandwich Theorem
• Average and Instantaneous Rate of Change
• Algebra of Derivative of Functions
• Product Rule in Derivatives
• Quotient Rule | Formula, Proof and Examples
• Derivatives of Polynomial Functions
• Application of Derivatives
• Applications of Power Rule

Chapter 14: Mathematical Reasoning

• Statements - Mathematical Reasoning
• Conditional Statements & Implications - Mathematical Reasoning | Class 11 Maths

Chapter 15: Statistics

• Measures of spread - Range, Variance, and Standard Deviation
• Mean Absolute Deviation
• Measures of Central Tendency in Statistics
• Difference Between Mean, Median, and Mode with Examples
• Frequency Distribution
• Variance and Standard Deviation

Chapter 16: Probability

• Random Experiment - Probability
• Types of Events in Probability
• Events in Probability
• Axiomatic Approach to Probability
• Measures of Dispersion | Types, Formula and Examples
• CBSE Class 11 Maths Formulas
• NCERT Solutions for Class 11 Maths - Chapter Wise PDF
• RD Sharma Class 11 Solutions for Maths

We are well versed with equations in multiple variables. Linear Equations represent a point in a single dimension, a line in a two-dimensional, and a plane in a three-dimensional world. Solutions to linear inequalities represent a region of the Cartesian plane. It becomes essential for us to know how to translate real-life problems into linear inequalities. 

Linear Inequalities 

Before defining the linear inequalities formally, let’s see them through a real-life situation and observe why their need arises in the first place. Let’s say Albert went to buy some novels for himself at the book fair. He has a total of Rs 200 with him. The book fair has a special sale policy which offers any book at Rs 70. Now he knows that he may not be able to spend the full amount on the books. Let’s say x is the number of books he bought. This situation can be represented mathematically by the following equation, 

70x < 200

Since he can’t spend all the amount on books, and also the amount spent by him will always be less than Rs 200. The present situation can only be represented by the equation given above. Now let’s study the linear inequalities with a formal description, 

Two real numbers or two algebraic expressions which are related by symbols such as ‘>’, ‘<‘, ‘≥’ and’≤’ form the inequalities. Linear inequalities are formed by linear equations which are connected with these symbols. These inequalities can be further classified into two parts:  Strict Inequalities: The inequalities with the symbols such as ‘>’ or ‘<‘. Slack Inequalities: The inequalities with the symbols such as ‘≥’ or ‘≤’.

Rules of Solving Linear Inequalities:

There are certain rules which we should keep in mind while solving linear inequalities. 

• Equal numbers can be added or subtracted from both sides of the inequality without affecting its sign.
• Both sides of Inequality can be divided or multiplied by any positive number but when they are multiplied or divided by a negative number, the sign of the linear inequality is reversed.

Now with this brief introduction to linear inequalities, let’s see some word problems on this concept. 

Sample Problems

Question 1: Considering the problem given in the beginning. Albert went to buy some novels for himself at the book fair. He has a total of Rs 200 with him. The book fair has a special sale policy which offers any book at Rs 70. Now he knows that he may not be able to spend the full amount on the books. Let’s say x is the number of books he bought. Represent this situation mathematically and graphically. 

Solution: 

We know that Albert cannot buy books for all the money he has. So, let’s say the number of books he buys is “x”. Then,  70x < 200 ⇒ x <  To plot the graph of this inequality, put x = 0.  0 <   Thus, x = 0 satisfies the inequality. So, the graph for the following inequality will look like, 

Question 2: Consider the performance of the strikers of the football club Real Madrid in the last 3 matches. Ronaldo and Benzema together scored less than 9 goals in the last three matches. It is also known that Ronaldo scored three more goals than Benzema. What can be the possible number of goals Ronaldo scored? 

Let’s say the number of goals scored by Benzema and Ronaldo are y and x respectively.  x = y + 3 …..(1) x + y < 9  …..(2)  Substituting the value of x from equation (1) in equation (2).  y + 3 + y < 9  ⇒2y  < 6  ⇒y < 3 Possible values of y: 0,1,2  Possible values of x: 3,4,5

Question 3: A classroom can fit at least 9 tables with an area of a one-meter square. We know that the perimeter of the classroom is 12m. Find the bounds on the length and breadth of the classroom. 

It can fit 9 tables, that means the area of the classroom is atleast 9m 2 . Let’s say the length of the classroom is x and breadth is y meters.  2(x + y) = 12 {Perimeter of the classroom} ⇒ x + y = 6  Area of the rectangle is given by,  xy > 9  ⇒x(6 – x) > 9  ⇒6x – x 2 > 9  ⇒ 0 > x 2 – 6x + 9  ⇒ 0 > (x – 3) 2 ⇒ 0 > x – 3  ⇒ x < 3  Thus, length of the classroom must be less than 3 m.  So, then the breadth of the classroom will be greater than 3 m. 

Question 4: Formulate the linear inequality for the following situation and plot its graph. 

Let’s say Aman and Akhil went to a stationery shop. Aman bought 3 notebooks and Akhil bought 4 books. Let’s say cost of each notebook was “x” and each book was “y”. The total expenditure was less than Rs 500. 

Cost of each notebook was “x” and for each book, it was “y”. Then the inequality can be described as, 3x + 4y < 500 Putting (x,y) → (0,0)  3(0) + 4(0) < 500 Origin satisfies the inequality. Thus, the graph of its solutions will look like, x.

Question 5: Formulate the linear inequality for the following situation and plot its graph. 

A music store sells its guitars at five times their cost price. Find the shopkeeper’s minimum cost price if his profit is more than Rs 3000. 

Let’s say the selling price of the guitar is y, and the cost price is x. y – x > 3000 ….(1) It is also given that,  y = 5x ….(2)  Substituting the value of y from equation (2) to equation (1).  5x – x > 3000  ⇒ 4x > 3000  ⇒  x >  ⇒ x> 750  Thus, the cost price must be greater than Rs 750. 

Question 6: The length of the rectangle is 4 times its breadth. The perimeter of the rectangle is less than 20. Formulate a linear inequality in two variables for the given situation, plot its graph and calculate the bounds for both length and breadth. 

Let’s say the length is “x” and breadth is “y”.  Perimeter = 2(x + y) < 20 ….(1) ⇒ x + y < 10  Given : x = 4y  Substituting the value of x in equation (1).  x + y < 10 ⇒ 5y < 10  ⇒ y < 2 So, x < 8 and y < 2. 

Question 7: Rahul and Rinkesh play in the same football team. In the previous game, Rahul scored 2 more goals than Rinkesh, but together they scored less than 8 goals. Solve the linear inequality and plot this on a graph.

The equations obtained from the given information in the question, Suppose Rahul scored x Number of goals and Rinkesh scored y number of goals, Equations obtained will be, x = y+2 ⇢ (1) x+y< 8 ⇢ (2) Solving both the equations, y+2 + y < 8 2y < 6 y< 3 Putting this value in equation (2), x< 5

Question 8: In a class of 100 students, there are more girls than boys, Can it be concluded that how many girls would be there?

Let’s suppose that B is denoted for boys and G is denoted for girls. Now, since Girls present in the class are more than boys, it can be written in equation form as, G > B  The total number of students present in class is 100 (given), It can be written as, G+ B= 100 B = 100- G Substitute G> B in the equation formed, G> 100 – G 2G > 100 G> 50 Hence, it is fixed that the number of girls has to be more than 50 in class, it can be 60, 65, etc. Basically any number greater than 50 and less than 100.

Question 9: In the previous question, is it possible for the number of girls to be exactly 50 or exactly 100? If No, then why?

No, It is not possible for the Number of girls to be exactly 50 since while solving, it was obtained that, G> 50  In any case if G= 50 is a possibility, from equation G+ B= 100, B = 50 will be obtained. This simply means that the number of boys is equal to the number of girls which contradicts to what is given in the question. No, it is not possible for G to be exactly 100 as well, as this proves that there are 0 boys in the class.

Question 10: Solve the linear inequality and plot the graph for the same,

7x+ 8y < 30

The linear inequality is given as, 7x+ 8y< 30  At x= 0, y= 30/8= 3.75 At y= 0, x= 30/7= 4.28 These values are the intercepts. The graph for the above shall look like,  Putting x= y/2, that is, y= 2x in the linear inequality, 7x + 16x < 30 x = 1.304 y = 2.609

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