Solving Inequality Word Questions

(You might like to read Introduction to Inequalities and Solving Inequalities first.)

In Algebra we have "inequality" questions like:

soccer teams

Sam and Alex play in the same soccer team. Last Saturday Alex scored 3 more goals than Sam, but together they scored less than 9 goals. What are the possible number of goals Alex scored?

How do we solve them?

The trick is to break the solution into two parts:

Turn the English into Algebra.

Then use Algebra to solve.

Turning English into Algebra

To turn the English into Algebra it helps to:

  • Read the whole thing first
  • Do a sketch if needed
  • Assign letters for the values
  • Find or work out formulas

We should also write down what is actually being asked for , so we know where we are going and when we have arrived!

The best way to learn this is by example, so let's try our first example:

Assign Letters:

  • the number of goals Alex scored: A
  • the number of goals Sam scored: S

We know that Alex scored 3 more goals than Sam did, so: A = S + 3

And we know that together they scored less than 9 goals: S + A < 9

We are being asked for how many goals Alex might have scored: A

Sam scored less than 3 goals, which means that Sam could have scored 0, 1 or 2 goals.

Alex scored 3 more goals than Sam did, so Alex could have scored 3, 4, or 5 goals .

  • When S = 0, then A = 3 and S + A = 3, and 3 < 9 is correct
  • When S = 1, then A = 4 and S + A = 5, and 5 < 9 is correct
  • When S = 2, then A = 5 and S + A = 7, and 7 < 9 is correct
  • (But when S = 3, then A = 6 and S + A = 9, and 9 < 9 is incorrect)

Lots More Examples!

pups

Example: Of 8 pups, there are more girls than boys. How many girl pups could there be?

  • the number of girls: g
  • the number of boys: b

We know that there are 8 pups, so: g + b = 8, which can be rearranged to

We also know there are more girls than boys, so:

We are being asked for the number of girl pups: g

So there could be 5, 6, 7 or 8 girl pups.

Could there be 8 girl pups? Then there would be no boys at all, and the question isn't clear on that point (sometimes questions are like that).

  • When g = 8, then b = 0 and g > b is correct (but is b = 0 allowed?)
  • When g = 7, then b = 1 and g > b is correct
  • When g = 6, then b = 2 and g > b is correct
  • When g = 5, then b = 3 and g > b is correct
  • (But if g = 4, then b = 4 and g > b is incorrect)

A speedy example:

bike

Example: Joe enters a race where he has to cycle and run. He cycles a distance of 25 km, and then runs for 20 km. His average running speed is half of his average cycling speed. Joe completes the race in less than 2½ hours, what can we say about his average speeds?

  • Average running speed: s
  • So average cycling speed: 2s
  • Speed = Distance Time
  • Which can be rearranged to: Time = Distance Speed

We are being asked for his average speeds: s and 2s

The race is divided into two parts:

  • Distance = 25 km
  • Average speed = 2s km/h
  • So Time = Distance Average Speed = 25 2s hours
  • Distance = 20 km
  • Average speed = s km/h
  • So Time = Distance Average Speed = 20 s hours

Joe completes the race in less than 2½ hours

  • The total time < 2½
  • 25 2s + 20 s < 2½

So his average speed running is greater than 13 km/h and his average speed cycling is greater than 26 km/h

In this example we get to use two inequalities at once:

ball throw

Example: The velocity v m/s of a ball thrown directly up in the air is given by v = 20 − 10t , where t is the time in seconds. At what times will the velocity be between 10 m/s and 15 m/s?

  • velocity in m/s: v
  • the time in seconds: t
  • v = 20 − 10t

We are being asked for the time t when v is between 5 and 15 m/s:

So the velocity is between 10 m/s and 15 m/s between 0.5 and 1 second after.

And a reasonably hard example to finish with:

Example: A rectangular room fits at least 7 tables that each have 1 square meter of surface area. The perimeter of the room is 16 m. What could the width and length of the room be?

Make a sketch: we don't know the size of the tables, only their area, they may fit perfectly or not!

  • the length of the room: L
  • the width of the room: W

The formula for the perimeter is 2(W + L) , and we know it is 16 m

  • 2(W + L) = 16
  • L = 8 − W

We also know the area of a rectangle is the width times the length: Area = W × L

And the area must be greater than or equal to 7:

  • W × L ≥ 7

We are being asked for the possible values of W and L

Let's solve:

So the width must be between 1 m and 7 m (inclusive) and the length is 8−width .

  • Say W = 1, then L = 8−1 = 7, and A = 1 x 7 = 7 m 2 (fits exactly 7 tables)
  • Say W = 0.9 (less than 1), then L = 7.1, and A = 0.9 x 7.1 = 6.39 m 2 (7 won't fit)
  • Say W = 1.1 (just above 1), then L = 6.9, and A = 1.1 x 6.9 = 7.59 m 2 (7 fit easily)
  • Likewise for W around 7 m

[FREE] Fun Math Games & Activities Packs

Always on the lookout for fun math games and activities in the classroom? Try our ready-to-go printable packs for students to complete independently or with a partner!

In order to access this I need to be confident with:

One step equations

Inequalities

Here you will learn about inequalities, including comparing quantities using inequalities, interpreting inequalities, representing inequalities, and solving inequalities.

Students first learn about inequality symbols in the first grade to represent relationships as part of their work with number and operations in base ten. They will expand upon this knowledge as they progress through elementary school into middle school, where they work with expressions and equations.

What are inequalities?

Inequalities are a comparison between two numbers, values, or expressions.

One of the quantities may be less than, greater than, less than or equal to, or greater than or equal to the other things.

This image describes the shape and direction of the symbols.

Inequalities image 1 US new image

When stacking identical blocks, the height of 3 blocks is greater than the height of 1 block. The lines joining the stacks give the shape and direction of the inequality symbols.

The symbols used to make the comparisons are in the table below.

\hspace{1.6cm} Comparison Symbols

Step by step guide: Great than signs

Step by step guide: Less than signs

  • You can use inequalities to make comparisons between quantities.
  • You can represent inequalities on a number line.

Recall, that x = 5 is an equation. The variable x is only equal to 5. In an equality, the value of x can be many numbers.

You can interpret and represent inequality statements on the number line.

An open circle shows it does not include the value.

A closed circle shows it does include the value.

  • You can solve simple inequalities.

What are inequalities?

Common Core State Standards

How does this relate to 1st through 6th grade math?

  • Grade 1 – Number and Operations Base 10 (1.NBT.B.3) Compare two two-digit numbers based on meanings of the tens and ones digits, recording the results of comparisons with the symbols >, \, =, and <.
  • Grade 2 – Number and Operations Base 10 (2.NBT.A.4) Compare two three-digit numbers based on meanings of the hundreds, tens, and ones digits, using >, \, =, and < symbols to record the results of comparisons.
  • Grade 3 – Number and Operations – Fractions (3.NF.A.3.d) Compare two fractions with the same numerator or the same denominator by reasoning about their size. Recognize that comparisons are valid only when the two fractions refer to the same whole. Record the results of comparisons with the symbols >, \, =, or <, and justify the conclusions, for example, by using a visual fraction model.
  • Grade 4 – Number and Operations Base Ten (4.NBT.A.2) Read and write multi-digit whole numbers using base-ten numerals, number names, and expanded form. Compare two multi-digit numbers based on meanings of the digits in each place, using >, \, =, and < symbols to record the results of comparisons.
  • Grade 4 – Number and Operations – Fractions (4.NF.A.2) Compare two fractions with different numerators and different denominators, for example by creating common denominators or numerators, or by comparing to a benchmark fraction such as \frac{1}{2}. Recognize that comparisons are valid only when the two fractions refer to the same whole. Record the results of comparisons with symbols >, \, =, or <, and justify the conclusions, for example, by using a visual fraction model.
  • Grade 4 – Number and Operations – Fractions (4.NF.C.7) Compare two decimals to hundredths by reasoning about their size. Recognize that comparisons are valid only when the two decimals refer to the same whole. Record the results of comparisons with the symbols >, \, =, or <, and justify the conclusions, for example, by using a visual model.
  • Grade 5 – Number and Operations Base Ten (5.NBT.A.3b) Compare two decimals to thousandths based on meanings of the digits in each place, using >, \, =, and < symbols to record the results of comparisons.
  • Grade 6 – Number System (6.NS.C.7.a) Interpret statements of inequality as statements about the relative position of two numbers on a number line diagram. For example, interpret –3 > –7 as a statement that –3 is located to the right of –7 on a number line oriented from left to right.
  • Grade 6 – Expressions and Equations (6.EE.B.8) Write an inequality of the form x > c or x < c to represent a constraint or condition in a real-world or mathematical problem. Recognize that inequalities of the form x > c or x < c have infinitely many solutions; represent solutions of such inequalities on number line diagrams.
  • Grade 6 – Expressions and Equations (6.EE.B.5) Understand solving an equation or inequality as a process of answering a question: which values from a specified set, if any, make the equation or inequality true? Use substitution to determine whether a given number in a specified set makes an equation or inequality true.
  • Grade 7 – Expressions and Equations (7.EE.4b) Solve word problems leading to inequalities of the form px + q > r or px + q < r, where p, q, and r are specific rational numbers. Graph the solution set of the inequality and interpret it in the context of the problem. For example: As a salesperson, you are paid \$50 per week plus \$3 per sale. This week you want your pay to be at least \$100. Write an inequality for the number of sales you need to make, and describe the solutions.

How to compare quantities using inequalities

In order to compare quantities using inequalities:

Choose a form to represent the quantities.

Compare the quantities.

  • Write the comparison using the appropriate inequality symbol

Step by step guide: Converting fractions, decimals and percentages

How to graph inequalities on a number line

In order to graph inequalities on a number line:

Identify the starting number.

Place a closed circle or open circle on the number.

  • Draw a line from the number in the direction representing the inequality.

How to solve inequalities

In order to solve one step inequalities:

Choose one side of the inequality to have the variable alone.

Use the additive inverse or multiplicative inverse to get the variable alone.

Write your solution with the inequality symbol.

Graph the solution set on the number line.

Step by step guide: Solving inequalities

[FREE] Inequalities Check for Understanding Quiz (Grade 1 to 7)

[FREE] Inequalities Check for Understanding Quiz (Grade 1 to 7)

Use this quiz to check your grade 1 to 7 students’ understanding of inequalities. 10+ questions with answers covering a range of 1st – 7th grade inequalities topics to identify areas of strength and support!

Inequalities examples

Example 1: compare fractions using an inequality.

Compare the fractions using a <, \, >, \, =.

Inequalities example 1

Both numbers are in the same form, so leave them as fractions.

Get a common denominator.

The common denominator between \, \cfrac{2}{3} \, and \, \cfrac{1}{4} is 12.

\begin{aligned} &\cfrac{2}{3}=\cfrac{2 \, \times \, 4}{3 \, \times \, 4}=\cfrac{8}{12}\\\\ &\cfrac{1}{4}=\cfrac{1 \, \times \, 3}{4 \, \times \, 3}=\cfrac{3}{12} \end{aligned}

2 Compare the quantities.

\cfrac{8}{12} \, is greater than \, \cfrac{3}{12}

3 Write the comparison using the appropriate inequality symbol.

\cfrac{8}{12} \, > \, \cfrac{3}{12}

So, \cfrac{2}{3} \, > \, \cfrac{1}{4}

For a detailed explanation, go to:

Step by step guide: Comparing Fractions

Example 2: compare percents and fractions using inequalities

Inequalities example 2

Change the percent to a fraction.

34 \%=\cfrac{34}{100}

Make \, \cfrac{2}{5} \, have a denominator of 100.

\cfrac{2}{5}=\cfrac{2 \, \times \, 20}{5 \, \times \, 20}=\cfrac{40}{100}

\cfrac{34}{100} \, is less than \, \cfrac{40}{100}

Write the comparison using the appropriate inequality symbol.

34 \% < \cfrac{2}{5}

Step by step guide: Converting Fractions, Decimals, and Percents

Example 3: represent an inequality on the number line

Represent x \geq-1

The starting value on the number line is -1.

Inequalities example 3

Draw a line from the starting number in the direction representing the inequality.

Inequalities example 3

The graph starts at -1 with a closed circle and the arrow is pointing to the right showing all the numbers greater than -1 or equal to -1.

Step by step guide: Inequalities on the number line

Step-by-step guide: Inequalities on a number line

Example 4: solve the inequality

Solve the inequality:

Keep the variable on the left side of the inequality symbol.

The additive inverse of +5 is -5 because + 5-5 = 0

\begin{aligned} & x+5 \leq-8 \\\\ & x+5-5 \leq-8-5 \\\\ & x \leq-13 \end{aligned}

Inequalities example 4

The solution set is all the numbers less than or equal to -13.

Example 5: solve the inequality

Solve the inequality: -4x < -16

Keep the variable on the left side of the inequality.

The multiplicative inverse of -4 is \cfrac{-1}{4}

\begin{aligned} & -4 x < -16 \\\\ & \cfrac{-4 x}{-4} \, < \, \cfrac{-16}{-4} \\\\ & x > 4 \end{aligned}

Change the direction of the inequality when dividing by -4.

Inequalities example 5

The solution set is all the numbers greater than 4.

Example 6: real world application of inequalities

Julie spends at least \$10.25 on each movie ticket. Write and graph an inequality representing this.

Identify the starting value.

The starting value on the number line is 10.25.

Decide if the starting value is included in the solution set.

10.25 will be in the solution set because the least amount of money Julie spends is \$10.25 , or she can spend more than that.

Identify the solution set with a straight line.

The inequality is: x ≥ 10.25

Inequalities example 6

The solution set is all the numbers greater than or equal to 10.25.

Teaching tips for inequalities

  • To help students memorize the inequality symbols, make sure they are posted in the classroom or somewhere else easily accessible. Also encourage students to read inequalities both forward and backwards, which forces them to focus on the shape of the inequality symbol.
  • When introducing the ≤ and ≥, explicitly make the connection between these signs and the greater than, less than, and equal sign.
  • Number line worksheets are a good way to help students develop this skill, but be sure to also include activities that are more interactive. This could include graphing inequalities digitally or using things that happen in and around the classroom as a basis for writing and graphing real world inequalities.
  • Demonstrate to students using the number line why the inequality symbol flips when multiplying or dividing by a negative number.
  • Reinforce conceptual understanding using a number line between inequalities and equalities.

Easy mistakes to make

  • Confusing the inequality symbols For example, comparing 5 < 4 and reading it as “5 is greater than 4” .

Inequalities image 5

Practice inequalities questions

1. Choose the correct symbol to make the comparison.

Inequalities practice 1

Compare the fractions by finding the common denominator and use equivalent fractions to change the fractions.

The common denominator is 14.

\begin{aligned} & \cfrac{2}{7}=\cfrac{2 \, \times \, 2}{7 \, \times \, 2}=\cfrac{4}{14} \\\\ & \cfrac{1}{2}=\cfrac{1 \, \times \, 7}{2 \, \times \, 7}=\cfrac{7}{14} \\\\ & \cfrac{4}{14} \, < \, \cfrac{7}{14} \end{aligned}

2. Choose the graph that represents the inequality.

Inequalities practice 2a

x \ge-1 is read as ‘x is greater than or equal to -1’

-1 is in the solution set so the circle is closed. The arrow points towards the right because that represents the numbers greater than.

Inequalities practice 2e

3. Solve the inequality:

Keep the variable on the left side of the inequality symbol. Use the additive inverse to solve the inequality.

The additive inverse of -7 is + 7 because -7 + 7 = 0. In other words, add 7 to both sides of the inequality.

\begin{aligned} & x-7<4 \\\\ & x-7+7<4+7 \\\\ & x<11 \end{aligned}

4. Solve the inequality:

5 x \geq 20

Keep the variable on the left side of the inequality symbol. Use the multiplicative inverse to solve the inequality.

The multiplicative inverse is 5 \times \cfrac{1}{5}=1. In other words, divide each side of the inequality by 5.

\begin{aligned} & 5 x>20 \\\\ & \cfrac{5 x}{5} > \cfrac{20}{5} \\\\ & x>4 \end{aligned}

5. Which number line represents the solution set to the following inequality?

x + 3 < 4

Inequalities practice 5a

The additive inverse of +3 is -3 because 3-3 = 0. In other words, subtract 3 from both sides of the inequality.

\begin{aligned} & x+3<4 \\\\ & x+3-3<4-3 \\\\ & x<1 \end{aligned}

6. Two times a number is greater than 6. Find the solution set.

Translate the problem into an inequality.

Solve the inequality. Keep the variable on the left side of the inequality symbol.

Use the multiplicative inverse to solve.

\begin{aligned} & 2 x>6 \\\\ & \cfrac{2 x}{2}>\cfrac{6}{2} \\\\ & x>3 \end{aligned}

Inequalities FAQs

Yes, when you get into pre-algebra and algebra you will be graphing compound inequalities on the number line which is more than one inequality.

Inequalities FAQS 1

No, just like expressions and equations, inequalities increasingly get more complex as students advance in their math learning. In 7th grade, students will continue learning by working with linear inequalities, also known as two-step inequalities.

The inequality symbol flips because you have to keep the inequality as a true comparison. For example, if you have 3 < 5 and you multiply both sides of the inequality by -1 you will get -3 < -5. This is no longer a true inequality comparison. In order to make it true, you will have to flip the inequality symbol, so < becomes >. -3 < -5 becomes -3 > -5.

Yes, the way you would solve a linear inequality follows the same process as you would do to solve a linear equation. With equations, you choose a side of the equation to bring the variable to, and you would do the same for an inequality. It’s most common to have the variable on the left hand side of an equality and the integers or real numbers on the right side.

Yes, when you get into high school math, you can use interval notation with parentheses to represent inequalities.

No, there are other types of inequalities, such as quadratic and polynomial inequalities and absolute value inequalities, that you will represent and solve in high school math.

The next lessons are

  • Types of graphs
  • Coordinate plane
  • Number patterns
  • Graphic inequalities
  • Linear inequalities
  • Quadratic inequalities

Still stuck?

At Third Space Learning, we specialize in helping teachers and school leaders to provide personalized math support for more of their students through high-quality, online one-on-one math tutoring delivered by subject experts.

Each week, our tutors support thousands of students who are at risk of not meeting their grade-level expectations, and help accelerate their progress and boost their confidence.

One on one math tuition

Find out how we can help your students achieve success with our math tutoring programs .

[FREE] Common Core Practice Tests (Grades 3 to 6)

Prepare for math tests in your state with these Grade 3 to Grade 6 practice assessments for Common Core and state equivalents.

40 multiple choice questions and detailed answers to support test prep, created by US math experts covering a range of topics!

Privacy Overview

inequality problem solving question

  • HW Guidelines
  • Study Skills Quiz
  • Find Local Tutors
  • Demo MathHelp.com
  • Join MathHelp.com

Select a Course Below

  • ACCUPLACER Math
  • Math Placement Test
  • PRAXIS Math
  • + more tests
  • 5th Grade Math
  • 6th Grade Math
  • Pre-Algebra
  • College Pre-Algebra
  • Introductory Algebra
  • Intermediate Algebra
  • College Algebra

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

MathHelp.com

Solving Linear Inequalities on MathHelp.com

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.

Advertisement

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% .

Algebra Tutors

  • 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.

URL: https://www.purplemath.com/modules/ineqlin3.htm

Page 1 Page 2 Page 3

Standardized Test Prep

College math, homeschool math, share this page.

  • Terms of Use
  • Privacy / Cookies
  • About Purplemath
  • About the Author
  • Tutoring from PM
  • Advertising
  • Linking to PM
  • Site licencing

Visit Our Profiles

inequality problem solving question

Download on App Store

  • Solve equations and inequalities
  • Simplify expressions
  • Factor polynomials
  • Graph equations and inequalities
  • Advanced solvers
  • All solvers
  • Arithmetics
  • Determinant
  • Percentages
  • Scientific Notation
  • Inequalities

Download on App Store

Equations and Inequalities Involving Signed Numbers

In chapter 2 we established rules for solving equations using the numbers of arithmetic. Now that we have learned the operations on signed numbers, we will use those same rules to solve equations that involve negative numbers. We will also study techniques for solving and graphing inequalities having one unknown.

SOLVING EQUATIONS INVOLVING SIGNED NUMBERS

Upon completing this section you should be able to solve equations involving signed numbers.

Example 1 Solve for x and check: x + 5 = 3

Using the same procedures learned in chapter 2, we subtract 5 from each side of the equation obtaining

inequality problem solving question

Example 2 Solve for x and check: - 3x = 12

Dividing each side by -3, we obtain

inequality problem solving question

LITERAL EQUATIONS

  • Identify a literal equation.
  • Apply previously learned rules to solve literal equations.

An equation having more than one letter is sometimes called a literal equation . It is occasionally necessary to solve such an equation for one of the letters in terms of the others. The step-by-step procedure discussed and used in chapter 2 is still valid after any grouping symbols are removed.

Example 1 Solve for c: 3(x + c) - 4y = 2x - 5c

First remove parentheses.

inequality problem solving question

At this point we note that since we are solving for c, we want to obtain c on one side and all other terms on the other side of the equation. Thus we obtain

inequality problem solving question

Sometimes the form of an answer can be changed. In this example we could multiply both numerator and denominator of the answer by (- l) (this does not change the value of the answer) and obtain

inequality problem solving question

The advantage of this last expression over the first is that there are not so many negative signs in the answer.

inequality problem solving question

The most commonly used literal expressions are formulas from geometry, physics, business, electronics, and so forth.

inequality problem solving question

Notice in this example that r was left on the right side and thus the computation was simpler. We can rewrite the answer another way if we wish.

inequality problem solving question

GRAPHING INEQUALITIES

  • Use the inequality symbol to represent the relative positions of two numbers on the number line.
  • Graph inequalities on the number line.

inequality problem solving question

The symbols are inequality symbols or order relations and are used to show the relative sizes of the values of two numbers. We usually read the symbol as "greater than." For instance, a > b is read as "a is greater than b." Notice that we have stated that we usually read a < b as a is less than b. But this is only because we read from left to right. In other words, "a is less than b" is the same as saying "b is greater than a." Actually then, we have one symbol that is written two ways only for convenience of reading. One way to remember the meaning of the symbol is that the pointed end is toward the lesser of the two numbers.

inequality problem solving question

In simpler words this definition states that a is less than b if we must add something to a to get b. Of course, the "something" must be positive.

If you think of the number line, you know that adding a positive number is equivalent to moving to the right on the number line. This gives rise to the following alternative definition, which may be easier to visualize.

Example 1 3 < 6, because 3 is to the left of 6 on the number line.

inequality problem solving question

Example 2 - 4 < 0, because -4 is to the left of 0 on the number line.

inequality problem solving question

Example 3 4 > - 2, because 4 is to the right of -2 on the number line.

inequality problem solving question

Example 4 - 6 < - 2, because -6 is to the left of -2 on the number line.

inequality problem solving question

The mathematical statement x < 3, read as "x is less than 3," indicates that the variable x can be any number less than (or to the left of) 3. Remember, we are considering the real numbers and not just integers, so do not think of the values of x for x < 3 as only 2, 1,0, - 1, and so on.

As a matter of fact, to name the number x that is the largest number less than 3 is an impossible task. It can be indicated on the number line, however. To do this we need a symbol to represent the meaning of a statement such as x < 3.

The symbols ( and ) used on the number line indicate that the endpoint is not included in the set.

Example 5 Graph x < 3 on the number line.

inequality problem solving question

Note that the graph has an arrow indicating that the line continues without end to the left.

Example 6 Graph x > 4 on the number line.

inequality problem solving question

Example 7 Graph x > -5 on the number line.

inequality problem solving question

Example 8 Make a number line graph showing that x > - 1 and x < 5. (The word "and" means that both conditions must apply.)

inequality problem solving question

Example 9 Graph - 3 < x < 3.

inequality problem solving question

Example 10 x >; 4 indicates the number 4 and all real numbers to the right of 4 on the number line.

The symbols [ and ] used on the number line indicate that the endpoint is included in the set.

inequality problem solving question

Example 13 Write an algebraic statement represented by the following graph.

inequality problem solving question

Example 14 Write an algebraic statement for the following graph.

inequality problem solving question

Example 15 Write an algebraic statement for the following graph.

inequality problem solving question

SOLVING INEQUALITIES

Upon completing this section you should be able to solve inequalities involving one unknown.

The solutions for inequalities generally involve the same basic rules as equations. There is one exception, which we will soon discover. The first rule, however, is similar to that used in solving equations.

If the same quantity is added to each side of an inequality , the results are unequal in the same order.

Example 1 If 5 < 8, then 5 + 2 < 8 + 2.

Example 2 If 7 < 10, then 7 - 3 < 10 - 3.

We can use this rule to solve certain inequalities.

Example 3 Solve for x: x + 6 < 10

If we add -6 to each side, we obtain

inequality problem solving question

Graphing this solution on the number line, we have

inequality problem solving question

We will now use the addition rule to illustrate an important concept concerning multiplication or division of inequalities.

Suppose x > a.

Now add - x to both sides by the addition rule.

inequality problem solving question

Now add -a to both sides.

inequality problem solving question

The last statement, - a > -x, can be rewritten as - x < -a. Therefore we can say, "If x > a, then - x < -a. This translates into the following rule:

If an inequality is multiplied or divided by a negative number, the results will be unequal in the opposite order.

Example 5 Solve for x and graph the solution: -2x>6

To obtain x on the left side we must divide each term by - 2. Notice that since we are dividing by a negative number, we must change the direction of the inequality.

inequality problem solving question

Take special note of this fact. Each time you divide or multiply by a negative number, you must change the direction of the inequality symbol. This is the only difference between solving equations and solving inequalities.

Once we have removed parentheses and have only individual terms in an expression, the procedure for finding a solution is almost like that in chapter 2.

Let us now review the step-by-step method from chapter 2 and note the difference when solving inequalities.

First Eliminate fractions by multiplying all terms by the least common denominator of all fractions. (No change when we are multiplying by a positive number.) Second Simplify by combining like terms on each side of the inequality. (No change) Third Add or subtract quantities to obtain the unknown on one side and the numbers on the other. (No change) Fourth Divide each term of the inequality by the coefficient of the unknown. If the coefficient is positive, the inequality will remain the same. If the coefficient is negative, the inequality will be reversed. (This is the important difference between equations and inequalities.)

inequality problem solving question

  • A literal equation is an equation involving more than one letter.
  • The symbols are inequality symbols or order relations .
  • a a is to the left of b on the real number line.
  • To solve a literal equation for one letter in terms of the others follow the same steps as in chapter 2.
  • To solve an inequality use the following steps: Step 1 Eliminate fractions by multiplying all terms by the least common denominator of all fractions. Step 2 Simplify by combining like terms on each side of the inequality. Step 3 Add or subtract quantities to obtain the unknown on one side and the numbers on the other. Step 4 Divide each term of the inequality by the coefficient of the unknown. If the coefficient is positive, the inequality will remain the same. If the coefficient is negative, the inequality will be reversed. Step 5 Check your answer.

Math Topics

More solvers.

  • Add Fractions
  • Simplify Fractions

Linear Inequalities Questions

Linear inequalities questions and solutions help the students to solve various problems on inequalities in mathematics. In this article, you will get solved questions on linear inequalities, and additional questions for practice. You will find detailed explanations to the solutions of linear inequalities in one variable and two variables.

What are Linear inequalities?

In maths, linear inequalities are the equations in which both sides of the equation are separated by the inequality symbol, such as <, >, ≤ and ≥. That means LHS and RHS are not equal. For example, 2x – 4 < 12, x – y > 7, 3x – 4 ≤ 0, and so on.

Also, check: Linear inequalities

Linear Inequalities Questions and Answers

1. Solve for x:

3(x – 1) ≤ 2 (x – 3)

The above inequality can be written as,

3x – 3 ≤ 2x – 6

Adding 3 to both the sides, we get;

3x – 3+ 3 ≤ 2x – 6+ 3

3x ≤ 2x – 3

Subtracting 2x from both the sides,

3x – 2x ≤ 2x – 3 – 2x

Therefore, the solutions to the given inequality are defined by all the real numbers less than or equal to -3.

Hence, the required solution set for x is (-∞, -3].

2. Solve the inequality: (x – 2)/(x + 5) > 2

(x – 2)/(x + 5) > 2

Subtracting 2 from both sides, we get;

(x – 2)/(x + 5) – 2 > 0

[(x – 2) – 2(x + 5)]/ (x + 5) > 0

(x – 2 – 2x – 10)/(x + 5) > 0

-(x + 12)/(x + 5) > 0

Multiplying -1 on both sides, we get;

(x + 12)/(x + 5) < 0

⇒ x + 12 < 0 and x + 5 > 0 (or) x + 12 > 0 and x + 5 < 0

⇒ x < -12 and x > -5 (or) x > -12 and x < -5

⇒ -12 < x < -5

Therefore, x ∈ (-12, -5).

3. Solve for x from the following:

1/(|x| – 3) ≤ ½

Subtracting ½ from both sides, we get;

[1/ (|x| – 3)] – (½) ≤ 0

(2 – |x| + 3)/ 2(|x| – 3) ≤ 0

(5 – |x|)/(|x| – 3) ≤ 0

⇒ 5 – |x| ≤ 0 and |x| – 3 > 0 or 5 – |x| ≥ 0 and |x| – 3 < 0

⇒ |x| ≥ 5 and |x| > 3 or |x| ≤ 5 and |x| < 3

⇒ |x| ≥ 5 or |x| < 3

⇒ x ∈ (- ∞ , – 5] or [5, ∞) or x ∈ ( -3 , 3)

⇒ x ∈ (- ∞ , – 5] ∪ ( -3 , 3) ∪ [5, ∞)

4. Solve for x: |x + 1| + |x| > 3

|x + 1| + |x| > 3

In LHS, we have two terms with modulus.

So, let’s equate the given expression within the modulus to 0.

Then, we get critical points x = -1, 0.

Thus, we can write the real line intervals for these critical points as: (-∞, -1), [-1, 0), [0, ∞)

When – ∞ < x < – 1

⇒ – x – 1 – x > 3

⇒ x < – 2.

When – 1 ≤ x < 0,

⇒ x + 1 – x > 3

⇒ 1 > 3 (not possible)

When 0 ≤ x < ∞,

⇒ x + 1 + x > 3

⇒ x > 1.

From the three cases written above, we get x ∈ (– ∞ , – 2) ∪ (1, ∞).

5. The longest side of a triangle is 3 times the shortest side, and the third side is 2 cm shorter than the longest side. If the perimeter of the triangle is at least 61 cm, find the minimum length of the shortest side.

Let x cm be the length of the shortest side of the triangle.

∴ According to the question, length of the longest side = 3x cm

Length of the third side = (3x – 2) cm

The least perimeter of the triangle = 61 cm (given)

Thus, x + 3x + (3x – 2) cm ≥ 61 cm

= 7x – 2 ≥ 61

Dividing by 7 on both sides, we get;

= 7x/7 ≥ 63/7

Hence, the minimum length of the shortest side will be 9 cm.

6. Solve (3x – 4)/2 ≥ (x + 1)/4 – 1. Show the graph of the solutions on the number line.

(3x – 4)/2 ≥ (x + 1)/4 – 1

(3x – 4)/2 ≥ [(x + 1) – 4]/4

(3x – 4)/2 ≥ (x – 3)/4

Multiplying by 4 on both sides,

(3x – 4) ≥ (x – 3)/2

2(3x – 4) ≥ (x – 3)

6x – 8 ≥ x – 3

Adding 8 on both sides,

Subtracting x from both sides, we get;

i.e., x ≥ 1

Therefore, the graphical representation of this solution (on the number line) is given as:

7. Solve the following system of linear inequalities graphically.

x + y ≥ 5….(i)

x – y ≤ 3….(ii)

From (i), x + y ≥ 5

Let x + y = 5

y = 5 – x

From (ii), x – y ≤ 3

Let x – y = 3

y = x – 3

Now, let’s draw the graph for the above two linear equations, i.e., x + y = 5 and x – y = 3.

Now, the solution of inequality (i) can be represented by shading the region towards the right of the line x + y = 5, including the points on the line.

On the same set of axes, we draw the graph of the equation x – y = 3. The inequality (ii) represents the shaded region towards the left side of the line x – y = 3, including the points on the line.

The double shaded region, common to the above two shaded regions, is the required solution region of the given system of inequalities.

8. A manufacturer has 600 litres of a 12% solution of acid. How many litres of a 30% acid solution must be added to it so that the acid content in the resulting mixture will be more than 15% but less than 18%?

Let x litres of 30% acid solution be required to be added.

Total mixture = (x + 600) litres

Thus, 30% x + 12% of 600 > 15% of (x + 600)

30% x + 12% of 600 < 18% of (x + 600)

⇒ (30x/100) + (12/100) × (600) > (15/100) (x + 600)

(30x/100) + (12/100) × (600) < (18/100) (x + 600)

⇒ 30x + 7200 > 15x + 9000 and 30x + 7200 < 18x + 10800

⇒ 15x > 1800 and 12x < 3600

⇒ x > 120 and x < 300,

i.e., 120 < x < 300

Hence, the number of litres of the 30% acid solution will have to be more than 120 litres but less than 300 litres.

9. Solve the following system of inequalities graphically.

x + y ≥ 4, 2x – y < 0

x + y ≥ 4….(i)

2x – y < 0….(ii)

From (i), x + y ≥ 4

Let x + y = 4

y = 4 – x

When x = 0, y = 4

When x = 4, y = 0

From (ii), 2x – y < 0

Let 2x – y = 0

When x = 0, y = 0

When x = 1, y = 2

From the above computation we can draw the graph of linear equations x + y = 4 and 2x – y = 0.

Now, shade the region for x + y ≥ 4, on the right of the line of x + y = 4.

For (ii), the shaded region will be on the left side of the line 2x – y = 0.

These can be represented graphically as:

Hence, the combined shaded region will be the solution region for the given system of inequalities.

10. Solve the following inequalities and represent the solution set on the number line.

5(2x – 7) – 3(2x + 3) ≤ 0, 2x + 19 ≤ 6x + 47

5(2x – 7) – 3(2x + 3) ≤ 0….(i)

2x + 19 ≤ 6x + 47….(ii)

5(2x – 7) – 3(2x + 3) ≤ 0

⇒ 10x – 35 – 6x – 9 ≤ 0

⇒ 4x – 44 ≤ 0

⇒ x ≤ 11 ……(iii)

2x + 19 ≤ 6x +47

⇒ 6x – 2x ≥ 19 – 47

⇒ x ≥ -7 ……….(iv)

From equations (iii) and (iv), the solution of the given inequalities is (-7, 11).

This can be shown on the number line as:

Practice Questions on Linear Inequalities

  • The cost and revenue functions of a product are given by C(x) = 20 x + 4000 and R(x) = 60x + 2000, respectively, where x is the number of items produced and sold. How many items must be sold to realise some profit?
  • Find all pairs of consecutive odd natural numbers, both of which are larger than 10, such that their sum is less than 40.
  • Solve the following inequalities and represent the solution set graphically on the number line.

5x + 1 > – 24, 5x – 1 < 24

  • Show that the following system of linear inequalities has no solution: x + 2y ≤ 3, 3x + 4y ≥ 12, x ≥ 0, y ≥ 1
  • Solve for x: -5 ≤ (5 – 3x)/2 ≤ 8

inequality problem solving question

  • Share Share

Register with BYJU'S & Download Free PDFs

Register with byju's & watch live videos.

close

  • Inequalities Practice Questions

Inequalities Practice Questions section is here now. The following section on Inequalities Practice Questions is complete and fully exhaustive. It covers most of the interesting topics and will let you perfect your problem-solving skills . The following Inequalities Practice Questions section has been divided into various sections for convenience. Let us begin our practice.

Inequalities Practice Questions – Type I

Directions : In the following questions the symbols #, *, @. $ and = are used with the following meanings:

A # B means A is greater than B.

A * B means A is greater than or equal to B.

A @ B means A is equal to B.

A $ B means A is lesser than B.

A = B means A is lesser than or equal to B.

Now in each of the following questions, assuming the three statements to be true, find which of the two conclusions I and II given below them is/are true. Give an answer.

A) if only conclusion I is true.

B) if only conclusion II is true

C) if either conclusion I or conclusion II is true

D) if neither conclusion I nor conclusion II is true

E) if both conclusions I and II are true.

Q1. Statements : P # Q, R $ P, R * O

Conclusions: I. Q # R II. Q $ R

Q2. Statements : P = Q, T @ R, R # P

Conclusions: I. T = Q II. Q *T

Q3. Statements : P @ Q, L @ M, P # L

Conclusions : I. Q # M II. M $ P

Q4. Statements : P # M # L, L# N @ Q, Q $ S @ R

Conclusions : I. R @ M II. L @ R

Q5. Statements : P * Q, Q@ T, T * L

Conclusions : I. Q # L II. T # P

Find Your Answers Here

Q1: D), Q2: C), Q3: E), Q4: D), Q5: D)

Inequalities Practice Questions

Directions: In the following questions the symbols @, @, =, © and © are used with the following meaning:

  • Here P @ Q means P is greater than Q.
  • P @ Q means P is either greater than or equal to Q.
  • P = Q means P is equal to Q.
  • The P © Q means P is smaller than Q.
  • P © Q means P is either smaller than or equal to Q.

Now in each of the following questions assuming the given statements to be true, find which of the two conclusions I and II given below them is/are definitely true? Give answer

A) if only conclusion I is true,

B) if only conclusion II is true,

C) if either I or II is true,

D) if neither 1 nor II is true, and

E) if both I and II are true.

Q1: Statements: B @ V K © C, C © B

Conclusions: I. V @ C II. B @ K

Q2: Statements: K @ T, $ = K,T © R

Conclusions: I. S @ R II. T = R

Q3: Statements: U = M, P @ U, M @ B

Conclusions: I. P = B II. P @ B

Q4: Statements : L @ N, J © P, P @ L

Conclusions: I. J = L II. P = N

Q5: Statements: H @ G, D @ E, H = E

Conclusions: I. D @ H II. G © D

Q1: B), Q2: D), Q3: C), Q4: D), Q5: E)

Directions: In the questions given below, certain symbols are used with the following meaning:

A @ B means A is greater than B.

A + B means A is either greater than or equal to B.

A # B means A is smaller than B

A ® B means A is either smaller than or equal to B.

A $ B means A is equal to B

Now in each of the following questions assuming the given statements to be true find which of the two conclusions I and II given below them is/are definitely true? Give answer

Q1: Statements : T $ G, K @ P, M # T, P + M

Conclusions: I. K @ T II. G $ P

Q2: Statements : R + N, S ® B, A @ , B $ A

Conclusions: I. S $ N II. A @ N

Q3: Statements : G $ K, F @ J, K + Q, Q + F

Conclusions: I. K $ F II. F # K

Q4: Statements : W @ S, K ® Z, U + W, S $ K

Conclusions: I. U @ K II. Z @ S

Q5: Statements: G $ E, D # K, E # S, K × G

Conclusions: I. S @ D II. D# E

Q1: D), Q2: B), Q3: C), Q4: A), Q5: E)

Directions: In the following questions the symbol $, @, *, ** and # are used with the following meaning.

A $ B means A is greater than B

A @ B means A is either greater than or equal to B

A * B means A is equal to B

A ** B means A is smaller than B

A # B means A is either smaller than or equal to B

Now in each of the following questions assuming the given statements to be true, find which of the two conclusions I and II given below them is/are definitely True ? Give answer

Q1: Statements: P @ Q, M # N, N**Q

Conclusions: I. P $ M II. N # P

Q2: Statements : D**X, F @ Y, D $ F

Conclusions : I. X @ Y II. Y # D

Q3: Statements : M**P, S $ T, M @ T

Conclusions : I. S * M II. T ** P

Q4: Statements : U*V, X $ W, U**W

Conclusions : I. W $ V II. U ** X

Q5: Statements : G $ H, J # K, H * K

Conclusions : I. H $ J II. J * H

Q1: A), Q2: D), Q3: B), Q4: E), Q5: C)

Customize your course in 30 seconds

Which class are you in.

tutor

Inequalities

  • Coded Inequalities
  • Word Problems
  • Quadratic Equations

Leave a Reply Cancel reply

Your email address will not be published. Required fields are marked *

Download the App

Google Play

inequality problem solving question

  • Testimonial
  • Web Stories

Hitbullseye Logo

Learning Home

inequality problem solving question

Not Now! Will rate later

inequality problem solving question

Inequalities : Practice Problems

inequality problem solving question

  • Rational & Modulus Inequalities: Theory & Concepts
  • Inequalities: Solved Examples
  • Higher order inequalities: Theory & Concepts
  • Inequalities: Practice Problems

inequality problem solving question

Most Popular Articles - PS

Time and Work Concepts

Time and Work Concepts

Time and Work Formula and Solved Problems

Time and Work Formula and Solved Problems

Time and Work Problems (Easy)

Time and Work Problems (Easy)

Time and Work Problems (Difficult)

Time and Work Problems (Difficult)

Problems on Ages Practice Problem : Level 02

Problems on Ages Practice Problems : Level 02

Chain Rule : Theory & Concepts

Chain Rule : Theory & Concepts

Chain Rule Solved Examples

Chain Rule Solved Examples

Chain Rule Practice Problems

Chain Rule Practice Problems: Level 01

Chain Rule Practice Problems

Chain Rule Practice Problems : Level 02

Problems on Numbers System : Level 02

Problems on Numbers System : Level 02

Download our app.

  • Learn on-the-go
  • Unlimited Prep Resources
  • Better Learning Experience
  • Personalized Guidance

Get More Out of Your Exam Preparation - Try Our App!

PrepScholar

Choose Your Test

Sat / act prep online guides and tips, inequalities on act math: strategies and practice.

feature_unbalance

Inequality questions come in a variety of shapes and forms on the ACT, but, no matter their form, you will see approximately three inequality questions on any given test . This means that inequality questions make up 5% of your overall ACT math test. Now, 5% of your test might not sound like a lot, but with only a quick brush-up on inequalities, that's an additional 5% of your questions that you're bound to rock!

This will be your complete guide to inequalities on the ACT : what they are, the different types of ACT math problems on inequalities, and how to solve them.

What Are Inequalities?

An inequality is a representation that two values are not equal or that two values are possibly not equal. There are different types of inequalities and different symbols to denote these different relationships.

≠ is the "unequal" sign. Whenever you see this sign, you know that two values are not equal, but nothing more. We don't know which value is greater or less than, just that they are not the same.

If we have $y ≠ x$, we do not know if $y$ is greater or less than $x$, just that they do not equal one another.

> is the "greater than" sign. Whichever number or variable is facing the opening of the sign is always the larger of the two values. (Some of you may have learned that the sign is a "crocodile" and that the crocodile always wants to eat the larger value).

For instance, $x > 14$ means that $x$ can be anything larger than 14 (it can even be 14.00000000001), but it cannot be 14 and it cannot be less than 14.

< is the "less than" sign. Whichever number is facing away from the opening of the sign is the lesser of the two values. This is just the greater than sign in reverse.

So $14 < x$ is the exact same equation we had earlier. $x$ must be larger than 14, 14 must be smaller than $x$.

≥ is the "greater than or equal to" sign. This acts exactly the same as the greater sign except for the fact that our values can also be equal.

Whereas $x > 14$ meant that $x$ could only be any number larger than 14, $x ≥ 14$ means that $x$ could be equal to 14 or could be any number larger than 14.

≤ is the "less than or equal to" sign. Just as the less than sign acted as a counter to the greater than sign, the less than or equal to sign acts counter to the greater than or equal to sign.

So $x ≥ 14$ is the exact same thing as saying $14 ≤ x$. Either way, we are saying that 14 is less than or equal to $x$, $x$ is greater than or equal to 14.

Each symbol describes the relationship between two values, but we can also link multiple values in a string. For instance, we can say:

$5 < x < 15$

This gives us both an upper and a lower limit on our $x$ value, because we know it must be both larger than five and less than 15. If we only had $5 < x$, the upper limit of $x$ would stretch into infinity, and the same with the lower limit if we only had $x < 15$.

For tips on how to keep track of which signs mean which, check out this article .

body_crocodile

The inequality crocodile is always hungry for the most it can get, om nom nom.

How to Represent Inequalities

We can represent inequalities in one of three different ways:

A written expression

A number line

Let's look at all three.

Inequalities as written expressions use only mathematical symbols and no diagrams. They are exactly what we have been working with above (e.g., $y > 37$).

An inequality number line allows us to visualize the set of numbers that represents our inequality. We use a dark line to show all the numbers that match our inequality, and we mark where the inequality begins and/or ends in two different ways. To mark the beginning of an inequality that is "greater than" or "less than," we use an open circle. This shows that the starting number is NOT included.

body_number_line_open

To mark the beginning of an inequality that is "greater than or equal to" or "less than or equal to," we use a closed circle. This shows that the starting number IS included.

body_number_line_closed

We can also combine these symbols if our inequality equation requires us to use two different symbols. For instance, if we have $-3 < x ≤ 3$, our number line would look like:

body_number_line_combo

And finally, we can have inequalities in graphs for any and all types of graphs on the coordinate plane ( more on the coordinate plane here ). "Greater than" will be above the line of the graph, while "less than" will be below the line of the graph.

body_inequality_graph_more

This is true no matter which direction the line of the graph extends.

body_inequality_graph_less

In terms of markings, inequality graphs follow the same rules as inequalities on number lines. Just as we use an open circle for "greater than" or "less than" inequalities, we use a dashed line for inequality graphs that are "greater than" or "less than."

body_graph_dash-1

And just how we use a closed circle for "greater than or equal to" or "less than or equal to" inequalities, we use a solid line for our graphs that are greater or less than or equal to.

body_graph_solid-1

Typical ACT Inequality Problems

There are three different types of inequality questions you'll see on the ACT, in the order from most to least common:

#1 : Solve an inequality equation (find the solution set)

#2 : Identify or answer questions about an inequality graph or number line

#3 : Find alternate inequalities that fulfill given information

Let's look at each type—what they mean and how you'll see them on the ACT.

#1: Solving an Inequality Equation

This is by far the most common type of inequality question you'll see on the ACT. You will be given one or two inequality equations and must solve for the solution set of your variable.

Inequality problems work exactly the same way as a single variable equation and can be solved in the same way. Just think of the inequality sign as being the same as the equals sign. So you will perform the same actions (adding, subtracting, multiplying, and dividing) on each side.

For instance:

$9 + 12x > 45$

$12x > 36$

The only difference between equations and inequalities is that the inequality sign flips if you multiply or divide each side by a negative .

For instance,

$10 - 4x < 50$

$-4x < 40$

$x > -10$

Because we had to divide each side by -4, we had to reverse the sign of the inequality.

Alternatively, we can also use the strategy of plugging in answers (PIA) or plugging in numbers (PIN) to solve our inequality problems. Because all ACT math problems are multiple choice, we can simply test out which answers match our equation (and which do not) or we can choose our own values for x based on the information we know, depending on the problem.

Let's look at an example of how this looks on the ACT, whether we solve the problem algebraically or by PIA.

The inequality $3(x+2)>4(x-3)$ is equivalent to which of the following inequalities?

F. $x<-6$ G. $x<5$ H. $x<9$ J. $x<14$ K. $x<18$

Solving Method 1: Algebra

First, distribute out the variable on each side.

$3(x + 2) > 4(x - 3)$

$3x + 6 > 4x - 12$

Now, we must isolate our variable just as we would with a single variable equation.

$6 > x - 12$

$18 > x$

Just as we saw back in our definitions, we know that we can also flip the inequality sign if we also switch the sides of our values. So $18 > x$ is the same as saying $x < 18$.

Our final answer is K , $x < 18$

Solving Method 2: Plugging in Answers

Though it will often take a little longer, we can also solve our inequality problems by testing out the values in our answer choices. Let's, as usual when using PIA, start with answer choice C.

Answer choice C says $x$ is less than 9, so let us see if this is true by saying that $x = 8$. If we plug in 8 for $x$ in the equation, we'll get:

$3(8 + 2) > 4(8 - 3)$

$3(10) > 4(5)$

$30 > 20$

This is true, but that doesn't necessarily mean that it is the correct answer. Just because we know that $x$ can be equal to 8 or less doesn't mean it can't also be greater than 8. All we know for sure is that we can eliminate answer choices F and G, since we've problem that $x$ can be larger than each of them. So let us now go the opposite route and look at the highest value $x$ can be, given our answer choices.

Answer choice J gives us $x < 14$ and answer choice K says that $x < 18$, so what would happen is we gave $x$ a value between the two? Let us say that $x = 16$

$3(16 + 2) > 4(16 - 3)$

$3(18) > 4(13)$

$54 > 52$

Because our inequality works for $x = 16$, we know that $x$ can be greater than $x < 14$ and can, therefore, be greater than all the answer choices except for answer choice K (the answer choice that gives us our largest possible value for $x$). This is enough to tell us that our final answer is K.

Our final answer is, again, K , $x < 18$

#2: Inequality Graph and Number Line Questions

For these types of questions, you will be asked to identify a graph or a number line from a given equation. Alternatively, you may be asked to infer information from a given inequality graph. Either way, you will always be given the graph on the coordinate plane.

body_ACT_inequalities_1_j

We know that the sum of $x$ and $y$ must be greater than 1, so let us imagine that one of those two variables is equal to 0.

If we say that $x = 0$, then y alone has to be greater than 1 to make the sum of $x$ and $y$ still be greater than 1. We also know that we indicate that a value is "greater than" on a graph with a dashed line at the value in question and a filled in area above the value.

The only graph with a dashed line at $y = 1$ and that has a shaded area above this value is graph J. This means graph J is more than likely our answer, but let's confirm it just to be safe.

Because the sum of $x$ and $y$ must be greater than 1, the alternative possibility to $x = 0$ and $y > 1$ is that $y$ equals zero, so $x$ must be greater than 1. To show this, we would need a dashed line at $(1, 0)$ and a shaded area above it, all of which graph J has.

Now, to finish confirming that graph J is indeed our answer, we would simply do what we did to locate the lower limit of our graph in reverse so that we can find the upper limit. If $x + y < 2$, then, when $x = 0$, $y$ must be less than 2, and when $y = 0$, $x$ must be less than 2. This would give us dashed lines at $(0, 2)$ and $(2, 0)$, both of which are on graph J.

Our final answer is J.

#3: Finding Alternate Inequality Expressions

The rarest form of inequality questions on the ACT will ask you to use given inequalities and find alternate inequalities that must be true based off this given information.

Let's look at one of these in action, to better see how this type of question works.

If $x$ and $y$ are real numbers, such that $x>1$ and $y<-1$, then which of the following inequalities must be true?

A. $x/y>1$

B. $|x|^2>|y|$

C. $x/3-5>y/3-5$

D. $x^2+1>y^2+1$

E. $x^{-2}>y^{-2}$

There are two different ways we can solve this problem, by plugging in our own numbers or by working through it based on our logic and knowledge of algebra. We'll go through both methods here.

Solving Method 1: Plugging in Numbers (PIN)

Because we have a problem with multiple variables in both the problem and in the answer choices, we can make life a little easier and give our variables numerical values. Now, we do have to be careful when using this method, however, because there are infinite variables to choose from for both $x$ and $y$ and so more than one answer choice might work for any given variables we give to $x$ and $y$. If two or more answer choices work, we must simply pick new variables—eventually only the correct answer will be left, as it must work for ALL values of $x$ and $y$.

When it comes to picking our values for $x$ and $y$, we can also make life easy by picking values that are easy to work with. We know that we must divide both $x$ and $y$ by 3 in answer choice C, so let us pick values that are divisible by 3, and we know we must square our values in several answer choices, so let us pick numbers that are fairly small.

Now let's just say that $x = 6$ and $y = -9$ (Why those numbers? So long as they fulfill the given information—and they do—then why not!)

And let us plug these values into our answer choices.

Answer choice A gives us:

$x/y > 1$

If we plug in our values, we get:

This is NOT greater than 1, so we can eliminate answer choice A.

Answer choice B gives us:

$|x|^2 > |y|$

$|6|^2 > |-9|$

$36 > 9$

This is correct, so we will keep answer option B in the running for right now.

Answer choice C gives us:

$x/3 - 5 > y/3 - 5$

$6/3 - 6 > {-9}/3 - 5$

$2 - 6 > -3 - 5$

$-4 > -8$

This is correct, so we will keep answer option C in the running for now as well. Because B and C are both correct, we will need to come back and test them both again with different values later.

Answer choice D gives us:

$x^2 + 1 > y^2 + 1$

$6^2 + 1 > -9^2 + 1$

$36 + 1 > 81 + 1$

$37 > 82$

This is NOT true, so we can eliminate answer choice D.

Answer choice E gives us:

$x^{-2} > y^{-2}$

$6^{-2} > -9^{-2}$

$1/{6^2} > 1/{-9^2}$

$1/36 > 1/81$

Now this is indeed true, but what if we had chosen different values for x and y? Let's say that we said $x = 9$ and $y = -6$ instead (remember—so long as the numbers fit with the given information, we can use any values we like).

$9^{-2} > -6^{-2}$

$1/{9^2} > 1/{-6^2}$

$1/81 > 1/36$

Whoops! Answer choice E is no longer correct, which means we can eliminate it. We are looking for the answer choice that is always true, so it cannot possibly be answer E.

Now we are left with answer choices B and C. Let's look at them each again.

While we saw that our values for $x$ and $y$ meant that answer choice B was indeed true, let's see what would happen if we choose a much smaller value for $y$. Nothing is stopping us from choosing -6,000 for $y$—remember, all that we are told is that $y < -1$. So let us use $y = -6,000$ instead.

$|6|^2 > |-6,000|$

$36 > 6,000$

This inequality is NOT true anymore, which means we can eliminate answer choice B.

This means that answer choice C must be the right answer by default, but let's test it to make absolutely sure.

Let us try what we did with answer option E and reverse the absolute values of our $x$ and $y$. So instead of $x = 6$ and $y = -9$, we will say that $x = 9$ and $y = -6$.

$9/3 - 5 > {-6}/3 - 5$

$3 - 5 > -2 - 5$

$-2 > -7$

No matter how many numbers we choose for $x$ and $y$, answer choice C will always be correct.

Our final answer is C , $x/3 - 5 > y/3 - 5$

Solving Method 2: Algebraic Logic

As we can see, using PIN was successful, but required a good deal of time and trial and error. The alternative way to solve the problem is by thinking of how negatives and positives work and how exponents and absolute values alter these rules.

We know that $x$ must be positive and $y$ must be negative to fulfill the requirements $x > 1$ and $y < -1$. Now let us look through our answer choices to see how these expressions are affected by the idea that $x$ must always be positive and $y$ must always be negative.

We know that any fraction with a positive numerator and a negative denominator will be negative. And any negative number is less than 1.

Answer choice A can never be correct.

An absolute value means that the negative sign on $y$ has been negated, so this might be correct. But y can be any number less than -1, which means its absolute value could potentially be astronomically large, and $x$ can be any number greater than 1, which means its absolute value might be comparatively tiny.

This means that answer choice B is not always correct, which is enough to eliminate it from the running.

Now let's look at each side of the inequality.

We know that any fraction with a positive number in both the numerator and in the denominator will give us a positive value. This means we will have some positive value minus 5 on the left side.

We also know that any time we have a negative value in the numerator and a positive value in the denominator, we will have a negative fraction. This means we will have some negative value minus 5 on the right side. We also know that a negative plus a negative will give us an even greater negative (a smaller value).

If we put this information together, we know that the left side may or may not be a negative value, depending on the value of $x$, but the right side will only get more and more negative. In other words, no matter what values we give to $x$ and $y$, the left side will always be greater than the right side, which means the expression is always true.

Now this should be enough for us to select our right answer as C, but we should give a look to the other answer choices just in case.

We know that if we square both a positive number and a negative number, we will get a positive result, so the negative value for $y$ is no longer in play. This inequality will therefore be true if the absolute value of $x$ is greater than the absolute value of $y$ (e.g., $x = 10$ and $y = -9$), but it won't be true if the absolute value of $y$ is greater than the absolute value of $x$ (e.g., $x = 9$, $y = -10$).

This means that the inequality will sometimes be true, but not always, which is enough to eliminate it.

Finally, answer choice E gives us:

We know that a number to a negative exponent is equal to 1 over that number to the positive exponent (e.g., $5^{-3} = 1/{5^3}$). This means that each value will be a fraction of 1 over the square of our $x$ and $y$ values.

This will give us two positive fractions and $1/{x^2}$ will only be larger if the absolute value of $x$ is smaller than the absolute value of $y$. But, because our $x$ and $y$ values can be anything so long as $y$ is negative and $x$ is positive, this will only sometimes be true. We can therefore eliminate answer choice E.

This leaves us with only answer choice C that is always true.

body_strategy-8

ACT Math Strategies for Inequality Problems

Though there are a few different types of inequality problems, there are a few strategies you can follow to help you solve them most effectively.

#1: Write Your Information Down and Draw It Out

Many problems on the ACT, inequalities included, appear easier or less complex than they actually are and can lead you to fall for bait answers. This illusion of ease may tempt you to try to solve inequality questions in your head, but this is NOT the way to go.

Take the extra moment to work your equations out on the paper or even draw your own diagrams (or draw on top of the diagrams you're given). The extra few seconds it will take you to write out your problems are well worth the points you'll gain by taking the time to find the right answer.

#2: Use PIN (or PIA ) When Necessary

If all you know about $x$ is that it must be more than 7, go ahead and pick a value for $\bi x$. This will help you more easily visualize and work through the rest of the problem, since it is generally always easier to work with numbers than it is to work with variables.

As you use this strategy, the safest bet is to choose two values for your variable—one that is close to the definition value and one that is very far away . This will allow you to see whether the values you chose work in all instances.

For instance, if all you know is $x > 7$, it's a good idea to work through the problem once under the assumption that $x = 8$ and another time under the assumption that $x = 400$. If the problem must be true for all values $x > 7$, then it should work for all numbers of $x$ greater than 7.

#3: Keep Very Careful Track of Your Negatives

One of the key differences between inequalities and single variable equations is in the fact that the inequality sign is reversed whenever you multiply or divide both sides by a negative . And you can bet the house that this is what the ACT will try to test you on again and again.

Though the ACT is not engineered to trick you , the test-makers are still trying to challenge you and test whether or not you know how to apply key mathematical concepts. If you lose track of your negatives (an easy thing to do, especially if you're working in your head), you will fall for one of the bait answers. Keep a keen eye.

#4: Double-Check Your Answer by Working Backwards (Optional)

If you feel unsure about your answer for any reason (because so many of the answer choices look the same, because you're not sure if you handled the issue of negative numbers correctly, etc.), you can work backwards to see if your expression is indeed correct.

For instance, let us look at the inequality we had earlier, when talking about the function of negatives on inequalities:

Again, we would go through this just as we would a single variable equation.

But now maybe that answer doesn't feel right to you or you just want to double-check to be sure. Well, if we're told that $x$ must be greater than -10 to fulfill the inequality, let's make sure that this is true.

Let us solve the expression with $x = -9$ and see if we are correct.

$10 - 4(-9) < 50$

$10 + 36 < 50$

$46 < 50$

This is correct, so that's promising. But we found that $x$ needed to be greater than -10, so our expression should also be INCORRECT if $x$ were equal to -10 or if $x$ were less than -10. So let us see what happens if we have $x = -10$.

$10 - 4(-10) < 50$

$10 + 40 < 50$

$50 < 50$

The inequality is no longer correct. This means that we know for certain that the solution set we found, $x > -10$ is true.

You will always be able to work backwards in this way to double-check your inequality questions. Though this can take a little extra time, it might be worth your peace of mind to do this whenever you feel unsure about your answers.

body_test-10

Ready, set? It's test time!

Test Your Knowledge

Now let's put all that inequality knowledge to the test on some real ACT math problems.

The inequality $6(x+2)>7(x-5)$ is equivalent to which of the following inequalities? A. $x<-23$ B. $x<7$ C. $x<17$ D. $x<37$ E. $x<47$

body_ACT_inequalities_6_e

If $r$ and $s$ can be any integers such that $s>10$ and $2r+s=15$, which of the following is the solution set for $r$?

A. $r≥3$ B. $r≥0$ C. $r≥2$ D. $r≤0$ E. $r≤2$

Which of the following is the solution statement for the inequality shown below?

$-5<1-3x<10$

F. $-5<x<10$ G. $-3<x$ H. $-3<x<2$ J. $-2<x<3$ K. $x<-3$ or $x>2$

body_ACT_inequalities_13_d

Answers: E, E, E, H, D

Answer Explanations

1. This is a standard inequality equation, so let us go through our solve accordingly. First, let's begin by distributing out our equation.

$6(x + 2) > 7(x - 5)$

$6x + 12 > 7x - 35$

$12 > x - 35$

$47 > x$

Because we did not have to divide or multiply by a negative, we were able to keep the inequality sign intact. And because the expression $47 > x$ and $x < 47$ mean the same thing, we can see that this matches one of our answer choices.

Our final answer is E , $x < 47$

2. We are given two graphs with equations attached and we must identify when one equation/graph is less than the other. We don't even have to know anything about what these equations means and we do not have to fuss with solving the equations—we can simply look at the diagram.

The only place on the diagram where the graph of $y = (x - 1)^4$ is less than (aka lower than) the graph of $y = x - 1$ is between $x = 1$ and $x = 2$ on the coordinate plane.

body_inequality_graph_colored

In other words, this inequality is true when $x > 1$ and when $x < 2$, or $1 < x < 2$.

Our final answer is E , $1 < x < 2$.

3. We know that $s > 10$ and it must be an integer, so let us make life easy and just say that $s = 11$. Now we can use this number to plug into the equation.

$2r + s = 15$

$2r + 11 = 15$

We know that $r$ can be equal to 2 and that it is the nearest integer to our definition. This means that our answer will either be C or E. So let us now find which direction our inequality sign must face.

Let's now try one integer larger than 11 to see whether our solution set must be less or equal to 2 or greater than or equal to 2. If we say that $s = 12$, then our equation becomes:

$2r + 12 = 15$

We can see now that, as $s$ increases, $r$ will decrease. This means our solution set will be that $r$ is equal to or less than 2.

Our final answer is E , $r≤ 2$

4. Though this problem is made slightly more complex due to the fact that it is a double inequality expression, we still solve the inequality the same way we normally would.

$-5 < 1 - 3x < 10$

If we think of this expression as two different inequality equations, we would say:

$-5 < 1 - 3x$

$1 - 3x < 10$

So let us solve each of them.

$-6 < -3x$

Because we now must divide by a negative, we must reverse the inequality sign.

And now let's solve our second expression:

$-3x < 9$

Again, we must reverse our inequality sign, since we need to divide each side by a negative.

$x > -3$

Now, if we put the two results together, our expression will be:

$-3 < x < 2$

Our final answer is H , $-3 < x < 2$

5. Because we have a number line with two closed circles, we know that must use less than or equal to and greater than or equal to signs.

We can see that the right side of the graph gives us a set of numbers equal to or greater than 3, which means:

The left side of the graph gives us a set of numbers less than or equal to -1, which means:

Our final answer is, therefore, D , $-1 ≥ x$ and $x ≤ 3$.

body_cute-1

The Take-Aways

Inequalities are so similar to single variable equations that it can be easy to treat the two as the same. The test-makers know this, so it pays to be careful when it comes to your inequality questions. Remember the key differences (multiplying or dividing by a negative reverses the sign, and you can flip your inequality signs so long as you flip both sides of the expression) and keep careful track of the details to avoid all the common pitfalls and bait answers.

After you've mastered the art of answering your inequality questions, that's another 5% of the test that you've dominated. You're well on your way to that score goal of yours now!

What's Next?

Want to brush up on any of your other math topics? Check out our individual math guides to get the walk-through on each and every topic on the ACT math test .

Been procrastinating on your ACT studying? Learn how to get over your desire to procrastinate and make a well-balanced study plan.

Running out of time on the ACT math section? We'll teach you how to beat the clock and maximize your ACT math score .

Looking to get a perfect score? Check out our guide to getting a perfect 36 on ACT math , written by a perfect-scorer.

Want to improve your ACT score by 4 points?

Check out our best-in-class online ACT prep program . We guarantee your money back if you don't improve your ACT score by 4 points or more.

Our program is entirely online, and it customizes what you study to your strengths and weaknesses . If you liked this Math lesson, you'll love our program. Along with more detailed lessons, you'll get thousands of practice problems organized by individual skills so you learn most effectively. We'll also give you a step-by-step program to follow so you'll never be confused about what to study next.

Check out our 5-day free trial:

Get 4 More Points on Your ACT, GUARANTEED

Courtney scored in the 99th percentile on the SAT in high school and went on to graduate from Stanford University with a degree in Cultural and Social Anthropology. She is passionate about bringing education and the tools to succeed to students from all backgrounds and walks of life, as she believes open education is one of the great societal equalizers. She has years of tutoring experience and writes creative works in her free time.

Student and Parent Forum

Our new student and parent forum, at ExpertHub.PrepScholar.com , allow you to interact with your peers and the PrepScholar staff. See how other students and parents are navigating high school, college, and the college admissions process. Ask questions; get answers.

Join the Conversation

Ask a Question Below

Have any questions about this article or other topics? Ask below and we'll reply!

Improve With Our Famous Guides

  • For All Students

The 5 Strategies You Must Be Using to Improve 160+ SAT Points

How to Get a Perfect 1600, by a Perfect Scorer

Series: How to Get 800 on Each SAT Section:

Score 800 on SAT Math

Score 800 on SAT Reading

Score 800 on SAT Writing

Series: How to Get to 600 on Each SAT Section:

Score 600 on SAT Math

Score 600 on SAT Reading

Score 600 on SAT Writing

Free Complete Official SAT Practice Tests

What SAT Target Score Should You Be Aiming For?

15 Strategies to Improve Your SAT Essay

The 5 Strategies You Must Be Using to Improve 4+ ACT Points

How to Get a Perfect 36 ACT, by a Perfect Scorer

Series: How to Get 36 on Each ACT Section:

36 on ACT English

36 on ACT Math

36 on ACT Reading

36 on ACT Science

Series: How to Get to 24 on Each ACT Section:

24 on ACT English

24 on ACT Math

24 on ACT Reading

24 on ACT Science

What ACT target score should you be aiming for?

ACT Vocabulary You Must Know

ACT Writing: 15 Tips to Raise Your Essay Score

How to Get Into Harvard and the Ivy League

How to Get a Perfect 4.0 GPA

How to Write an Amazing College Essay

What Exactly Are Colleges Looking For?

Is the ACT easier than the SAT? A Comprehensive Guide

Should you retake your SAT or ACT?

When should you take the SAT or ACT?

Stay Informed

inequality problem solving question

Get the latest articles and test prep tips!

Looking for Graduate School Test Prep?

Check out our top-rated graduate blogs here:

GRE Online Prep Blog

GMAT Online Prep Blog

TOEFL Online Prep Blog

Holly R. "I am absolutely overjoyed and cannot thank you enough for helping me!”

Library homepage

  • school Campus Bookshelves
  • menu_book Bookshelves
  • perm_media Learning Objects
  • login Login
  • how_to_reg Request Instructor Account
  • hub Instructor Commons
  • Download Page (PDF)
  • Download Full Book (PDF)
  • Periodic Table
  • Physics Constants
  • Scientific Calculator
  • Reference & Cite
  • Tools expand_more
  • Readability

selected template will load here

This action is not available.

Mathematics LibreTexts

3.7: Solving Systems of Inequalities with Two Variables

  • Last updated
  • Save as PDF
  • Page ID 6400

Learning Objectives

  • Check solutions to systems of inequalities with two variables.
  • Graph solution sets of systems of inequalities.

Solutions to Systems of Inequalities

A system of inequalities 33 consists of a set of two or more inequalities with the same variables. The inequalities define the conditions that are to be considered simultaneously. For example,

\(\left\{ \begin{array} { l } { y > x - 2 } \\ { y \leq 2 x + 2 } \end{array} \right.\)

We know that each inequality in the set contains infinitely many ordered pair solutions defined by a region in a rectangular coordinate plane. When considering two of these inequalities together, the intersection of these sets will define the set of simultaneous ordered pair solutions. When we graph each of the above inequalities separately we have:

39339dc174d251b4a24618b8472b6556.png

And when graphed on the same set of axes, the intersection can be determined.

84a446b513683b620bb90e8d8248e74c.png

The intersection is shaded darker and the final graph of the solution set will be presented as follows:

47565cb2a79e6b946bac644d970ac382.png

The graph suggests that \((3, 2)\) is a solution because it is in the intersection. To verify this, we can show that it solves both of the original inequalities as follows:

Points on the solid boundary are included in the set of simultaneous solutions and points on the dashed boundary are not. Consider the point \((−1, 0)\) on the solid boundary defined by \(y = 2x + 2\) and verify that it solves the original system:

Notice that this point satisfies both inequalities and thus is included in the solution set. Now consider the point \((2, 0)\) on the dashed boundary defined by \(y = x − 2\) and verify that it does not solve the original system:

This point does not satisfy both inequalities and thus is not included in the solution set.

Example \(\PageIndex{1}\)

Determine whether or not \((-3,3)\) is a solution to the following system:

\(\left\{ \begin{aligned} 2 x + 6 y \leq 6 \\ - \frac { 1 } { 3 } x - y \leq 3 \end{aligned} \right.\)

Substitute the coordinates of \((x, y) = (−3, 3)\) into both inequalities.

\((-3,3)\) is not a solution; it does not satisfy both inequalities.

We can graph the solutions of systems that contain nonlinear inequalities in a similar manner. For example, both solution sets of the following inequalities can be graphed on the same set of axes:

\(\left\{ \begin{array} { l } { y < \frac { 1 } { 2 } x + 4 } \\ { y \geq x ^ { 2 } } \end{array} \right.\)

02cff5ff90cd08aa1717d8e8f6f1b2cd.png

And the intersection of both regions contains the region of simultaneous ordered pair solutions.

4d25116eb893d726af7f80bd3137a36e.png

From the graph, we expect the ordered pair \((1, 3)\) to solve both inequalities.

Graphing Solutions to Systems of Inequalities

Solutions to a system of inequalities are the ordered pairs that solve all the inequalities in the system. Therefore, to solve these systems we graph the solution sets of the inequalities on the same set of axes and determine where they intersect. This intersection, or overlap, will define the region of common ordered pair solutions.

Example \(\PageIndex{2}\):

Graph the solution set: \(\left\{ \begin{array} { l } { - 2 x + y > - 4 } \\ { 3 x - 6 y \geq 6 } \end{array} \right.\).

To facilitate the graphing process, we first solve for \(y\).

\(\left\{ \begin{array} { l l } { - 2 x + y > - 4 } \\ { 3 x - 6 y \geq 6 } \end{array} \right. \quad\Rightarrow\quad \left\{ \begin{array} { l } { y > 2 x - 4 } \\ { y \leq \frac { 1 } { 2 } x - 1 } \end{array} \right.\)

For the first inequality, we use a dashed boundary defined by \(y = 2x − 4\) and shade all points above the line. For the second inequality, we use a solid boundary defined by \(y = \frac{1}{ 2} x − 1\) and shade all points below. The intersection is darkened.

1e345dc5e6d3613c331276587ebe345c.png

Now we present our solution with only the intersection shaded.

f78b4540cde6e88dd3f509f538b708f1.png

Example \(\PageIndex{3}\):

Graph the solution set: \(\left\{ \begin{array} { l } { - 3 x + 2 y > 6 } \\ { 6 x - 4 y > 8 } \end{array} \right.\).

We begin by solving both inequalities for \(y\).

\(\left\{ \begin{array} { l } { - 3 x + 2 y > 6 } \\ { 6 x - 4 y > 8 } \end{array} \right. \quad\Rightarrow\quad \left\{ \begin{array} { l } { y > \frac { 3 } { 2 } x + 3 } \\ { y < \frac { 3 } { 2 } x - 2 } \end{array} \right.\)

Because of the strict inequalities, we will use a dashed line for each boundary. For the first inequality shade all points above the boundary and for the second inequality shade all points below the boundary.

0b3f328e7ef8eb42e2abc3c8855b17ea.png

As we can see, there is no intersection of these two shaded regions. Therefore, there are no simultaneous solutions.

\(\varnothing\)

Example \(\PageIndex{4}\):

Graph the solution set: \(\left\{ \begin{array} { l } { y \geq - 4 } \\ { y < x + 3 } \\ { y \leq - 3 x + 3 } \end{array} \right.\)

Begin by graphing the solution sets to all three inequalities.

9fe4aa3a78b7a9889bb5273079d91f26.png

After graphing all three inequalities on the same set of axes, we determine that the intersection lies in the triangular region pictured below.

f45683770f75ba64090316ac224dded5.png

The graph suggests that \((−1, 1)\) is a simultaneous solution. As a check, we could substitute that point into the inequalities and verify that it solves all three conditions.

Use the same technique to graph the solution sets to systems of nonlinear inequalities.

Example \(\PageIndex{5}\):

Graph the solution set: \(\left\{ \begin{array} { l } { y < ( x + 1 ) ^ { 2 } } \\ { y \leq - \frac { 1 } { 2 } x + 3 } \end{array} \right.\).

The first inequality has a parabolic boundary. This boundary is a horizontal translation of the basic function \(y = x^{2}\) to the left \(1\) unit. Because of the strict inequality, the boundary is dashed, indicating that it is not included in the solution set. The second inequality is linear and will be graphed with a solid boundary. Solution sets to both are graphed below.

a77d47a78c2d0c2f35d78f4a2904c206.png

After graphing the inequalities on the same set of axes, we determine that the intersection lies in the region pictured below.

c1ea1886c32af362fb3cddd88a45fe62.png

Exercise \(\PageIndex{1}\)

Graph the solution set: \(\left\{ \begin{array} { l } { y \geq - | x + 1 | + 3 } \\ { y \leq 2 } \end{array} \right.\).

5906eac1da9f9dd2ddcfa5b88d2fde17.png

www.youtube.com/v/9z87e7Iw9JE

Key Takeaways

  • To graph solutions to systems of inequalities, graph the solution sets of each inequality on the same set of axes and determine where they intersect.
  • You can check your answer by choosing a few values inside and out of the shaded region to see if they satisfy the inequalities or not. While this is not a proof, doing so will give a good indication that you have graphed the correct region.

Exercise \(\PageIndex{2}\)

Determine whether or not the given point is a solution to the given system of inequalities.

1. \((-2,1)\);

\(\left\{ \begin{array} { l } { y > 3 x + 5 } \\ { y \leq - x + 1 } \end{array} \right.\)

2. \((-1,-3)\);

\(\left\{ \begin{array} { l } { y \geq 3 x - 1 } \\ { y < - 2 x } \end{array} \right.\)

3. \((-2,-1)\);

\(\left\{ \begin{array} { l } { x - 2 y > - 1 } \\ { 3 x - y < - 3 } \end{array} \right.\)

4. \((0,-5)\);

\(\left\{ \begin{array} { c } { 5 x - y \geq 5 } \\ { 3 x + 2 y < - 1 } \end{array} \right.\)

5. \((-\frac{1}{2} ,0)\);

\(\left\{ \begin{array} { l } { - 8 x + 5 y \geq 3 } \\ { 2 x - 3 y < 0 } \end{array} \right.\)

6. \((-1, \frac{1}{2})\);

\(\left\{ \begin{array} { l } { 2 x - 9 y < - 1 } \\ { 3 x - 6 y > - 2 } \end{array} \right.\)

7. \((-1,-2)\);

\(\left\{ \begin{array} { c } { 2 x - y \geq - 1 } \\ { x - 3 y < 6 } \\ { 2 x - 3 y > - 1 } \end{array} \right.\)

8. \((-5,2)\);

\(\left\{ \begin{array} { c } { - x + 5 y > 10 } \\ { 2 x + y < 1 } \\ { x + 3 y < - 2 } \end{array} \right.\)

9. \((0,3)\);

\(\left\{ \begin{array} { l } { y + 4 \geq 0 } \\ { \frac { 1 } { 2 } x + \frac { 1 } { 3 } y \leq 1 } \\ { - 3 x + 2 y \leq 6 } \end{array} \right.\)

10. \((1,1)\);

  • \(\left\{ \begin{array} { l } { y \leq - \frac { 3 } { 4 } x + 2 } \\ { y \geq - 5 x + 2 } \\ { y \geq \frac { 1 } { 3 } x - 1 } \end{array} \right.\)

11. \((-1,2)\);

\( \left\{ \begin{array} { l } { y \geq x ^ { 2 } + 1 } \\ { y < - 2 x + 3 } \end{array} \right.\)

12. \((4,5)\);

\(\left\{ \begin{array} { l } { y < ( x - 1 ) ^ { 2 } - 1 } \\ { y > \frac { 1 } { 2 } x - 1 } \end{array} \right.\)

13. \((-2,-3)\);

\(\left\{ \begin{array} { l } { y < 0 } \\ { y \geq - | x | + 4 } \end{array} \right.\)

14. \((1,2)\);

\(\left\{ \begin{array} { l } { y < | x - 3 | + 2 } \\ { y \geq 2 } \end{array} \right.\)

15. \(\left( - \frac { 1 } { 2 } , - 5 \right)\);

\(\left\{ \begin{array} { l } { y \leq - 3 x - 5 } \\ { y > ( x - 1 ) ^ { 2 } - 10 } \end{array} \right.\)

16. \((-4,1)\)

\(\left\{ \begin{array} { l } { x \geq - 5 } \\ { y < ( x + 3 ) ^ { 2 } - 2 } \end{array} \right.\)

17. \(\left( - \frac { 3 } { 2 } , \frac { 1 } { 3 } \right)\);

\(\left\{ \begin{array} { l } { x - 2 y \leq 4 } \\ { y \leq | 3 x - 1 | + 2 } \end{array} \right.\)

18, \(\left( - 3 , - \frac { 3 } { 4 } \right)\);

\(\left\{ \begin{array} { l } { 3 x - 4 y < 24 } \\ { y < ( x + 2 ) ^ { 2 } - 1 } \end{array} \right.\)

19. \((4,2)\);

  • \(\left\{ \begin{array} { l } { y < ( x - 3 ) ^ { 2 } + 1 } \\ { y < - \frac { 3 } { 4 } x + 5 } \end{array} \right.\)

20. \((\frac{5}{2}, 1)\)

  • \(\left\{ \begin{array} { l } { y \geq - 1 } \\ { y < - ( x - 2 ) ^ { 2 } + 3 } \end{array} \right.\)

Exercise \(\PageIndex{3}\)

Graph the solution set.

  • \(\left\{ \begin{array} { l } { y \geq \frac { 2 } { 3 } x - 3 } \\ { y < - \frac { 1 } { 3 } x + 3 } \end{array} \right.\)
  • \(\left\{ \begin{array} { l } { y \geq - \frac { 1 } { 4 } x + 1 } \\ { y < \frac { 1 } { 2 } x - 2 } \end{array} \right.\)
  • \(\left\{ \begin{array} { l } { y > \frac { 2 } { 3 } x + 1 } \\ { y > \frac { 4 } { 3 } x - 5 } \end{array} \right.\)
  • \(\left\{ \begin{array} { l } { y \leq - 5 x + 4 } \\ { y < \frac { 4 } { 3 } x - 2 } \end{array} \right.\)
  • \(\left\{ \begin{array} { l } { x - y \geq - 3 } \\ { x + y \geq 3 } \end{array} \right.\)
  • \(\left\{ \begin{array} { l } { 3 x + y < 4 } \\ { 2 x - y \leq 1 } \end{array} \right.\)
  • \(\left\{ \begin{array} { l } { - x + 2 y \leq 0 } \\ { 3 x + 5 y < 15 } \end{array} \right.\)
  • \(\left\{ \begin{array} { c } { 2 x + 3 y < 6 } \\ { - 4 x + 3 y \geq - 12 } \end{array} \right.\)
  • \(\left\{ \begin{array} { l } { 3 x + 2 y > 1 } \\ { 4 x - 2 y > 3 } \end{array} \right.\)
  • \(\left\{ \begin{array} { l } { x - 4 y \geq 2 } \\ { 8 x + 4 y \leq 3 } \end{array} \right.\)
  • \(\left\{ \begin{array} { l } { 5 x - 2 y \leq 6 } \\ { - 5 x + 2 y < 2 } \end{array} \right.\)
  • \(\left\{ \begin{array} { l } { 12 x + 10 y > 20 } \\ { 18 x + 15 y < - 15 } \end{array} \right.\)
  • \(\left\{ \begin{array} { l } { x + y < 0 } \\ { y + 4 > 0 } \end{array} \right.\)
  • \(\left\{ \begin{array} { l } { x > - 3 } \\ { y < 1 } \end{array} \right.\)
  • \(\left\{ \begin{array} { l } { 2 x - 2 y < 0 } \\ { 3 x - 3 y > 3 } \end{array} \right.\)
  • \(\left\{ \begin{array} { l } { y + 1 \leq 0 } \\ { y + 3 \geq 0 } \end{array} \right.\)
  • Construct a system of linear inequalities that describes all points in the first quadrant.
  • Construct a system of linear inequalities that describes all points in the second quadrant.
  • Construct a system of linear inequalities that describes all points in the third quadrant.
  • Construct a system of linear inequalities that describes all points in the fourth quadrant.

a3d39b3bc22bd7937852dc87ef7f5393.png

15. \(\varnothing\)

17. \(\left\{ \begin{array} { l } { x > 0 } \\ { y > 0 } \end{array} \right.\)

19. \(\left\{ \begin{array} { l } { x < 0 } \\ { y < 0 } \end{array} \right.\)

Exercise \(\PageIndex{4}\)

  • \(\left\{ \begin{array} { l } { y \geq - \frac { 1 } { 2 } x + 3 } \\ { y \geq \frac { 3 } { 2 } x - 3 } \\ { y \leq \frac { 3 } { 2 } x + 1 } \end{array} \right.\)
  • \(\left\{ \begin{array} { l } { 3 x - 2 y > 6 } \\ { 5 x + 2 y > 8 } \\ { - 3 x + 4 y \leq 4 } \end{array} \right.\)
  • \(\left\{ \begin{array} { l } { 3 x - 5 y > - 15 } \\ { 5 x - 2 y \leq 8 } \\ { x + y < - 1 } \end{array} \right.\)
  • \(\left\{ \begin{array} { l } { 3 x - 2 y < - 1 } \\ { 5 x + 2 y > 7 } \\ { y + 1 > 0 } \end{array} \right.\)
  • \(\left\{ \begin{array} { l } { 3 x - 2 y < - 1 } \\ { 5 x + 2 y < 7 } \\ { y + 1 > 0 } \end{array} \right.\)
  • \(\left\{ \begin{array} { l } { 4 x + 5 y - 8 < 0 } \\ { y > 0 } \\ { x + 3 > 0 } \end{array} \right.\)
  • \(\left\{ \begin{array} { l } { y - 2 < 0 } \\ { y + 2 > 0 } \\ { 2 x - y \geq 0 } \end{array} \right.\)
  • \(\left\{ \begin{array} { l } { \frac { 1 } { 2 } x + \frac { 1 } { 2 } y < 1 } \\ { x < 3 } \\ { - \frac { 1 } { 2 } x + \frac { 1 } { 2 } y \leq 1 } \end{array} \right.\)
  • \(\left\{ \begin{array} { l } { \frac { 1 } { 2 } x + \frac { 1 } { 3 } y \leq 1 } \\ { y + 4 \geq 0 } \\ { - \frac { 1 } { 2 } x + \frac { 1 } { 3 } y \leq 1 } \end{array} \right.\)
  • \(\left\{ \begin{array} { l } { y < x + 2 } \\ { y \geq x ^ { 2 } - 3 } \end{array} \right.\)
  • \(\left\{ \begin{array} { l } { y \geq x ^ { 2 } + 1 } \\ { y > - \frac { 3 } { 4 } x + 3 } \end{array} \right.\)
  • \(\left\{ \begin{array} { l } { y \leq ( x + 2 ) ^ { 2 } } \\ { y \leq \frac { 1 } { 3 } x + 4 } \end{array} \right.\)
  • \(\left\{ \begin{array} { l } { y < - ( x + 1 ) ^ { 2 } - 1 } \\ { y < \frac { 3 } { 2 } x - 2 } \end{array} \right.\)
  • \(\left\{ \begin{array} { l } { y \leq \frac { 1 } { 3 } x + 3 } \\ { y \geq | x + 3 | - 2 } \end{array} \right.\)
  • \(\left\{ \begin{array} { l } { y \leq - x + 5 } \\ { y > | x - 1 | + 2 } \end{array} \right.\)
  • \(\left\{ \begin{array} { l } { y > - | x - 2 | + 5 } \\ { y > 2 } \end{array} \right.\)
  • \(\left\{ \begin{array} { l } { y \leq - | x | + 3 } \\ { y < \frac { 1 } { 4 } x } \end{array} \right.\)
  • \(\left\{ \begin{array} { l } { y > | x | + 1 } \\ { y \leq x - 1 } \end{array} \right.\)
  • \(\left\{ \begin{array} { l } { y \leq | x | + 1 } \\ { y > x - 1 } \end{array} \right.\)
  • \(\left\{ \begin{array} { l } { y \leq | x - 3 | + 1 } \\ { x \leq 2 } \end{array} \right.\)
  • \(\left\{ \begin{array} { l } { y > | x + 1 | } \\ { y < x - 2 } \end{array} \right.\)
  • \(\left\{ \begin{array} { l } { y < x ^ { 3 } + 2 } \\ { y \leq x + 3 } \end{array} \right.\)
  • \(\left\{ \begin{array} { l } { y \leq 4 } \\ { y \geq ( x + 3 ) ^ { 3 } + 1 } \end{array} \right.\)
  • \(\left\{ \begin{array} { l } { y \geq - 2 x + 6 } \\ { y > \sqrt { x } + 3 } \end{array} \right.\)
  • \(\left\{ \begin{array} { l } { y \leq \sqrt { x + 4 } } \\ { x \leq - 1 } \end{array} \right.\)
  • \(\left\{ \begin{array} { l } { y \leq - x ^ { 2 } + 4 } \\ { y \geq x ^ { 2 } - 4 } \end{array} \right.\)
  • \(\left\{ \begin{array} { l } { y \geq | x - 1 | - 3 } \\ { y \leq - | x - 1 | + 3 } \end{array} \right.\)

4ba0469110372d99eb1b2725d6daea30.png

21. \(\varnothing\)

60d61f885868e71dc08882facdf107e9.png

33 A set of two or more inequalities with the same variables.

If you're seeing this message, it means we're having trouble loading external resources on our website.

If you're behind a web filter, please make sure that the domains *.kastatic.org and *.kasandbox.org are unblocked.

To log in and use all the features of Khan Academy, please enable JavaScript in your browser.

Unit 3: Linear equations and inequalities

About this unit, one-step equations.

  • One-step addition & subtraction equations (Opens a modal)
  • One-step addition equation (Opens a modal)
  • One-step division equations (Opens a modal)
  • One-step multiplication equations (Opens a modal)
  • One-step multiplication & division equations (Opens a modal)
  • One-step addition & subtraction equations Get 5 of 7 questions to level up!
  • One-step multiplication & division equations Get 5 of 7 questions to level up!

Two-steps equations

  • Intro to two-step equations (Opens a modal)
  • Two-step equations intuition (Opens a modal)
  • Worked example: two-step equations (Opens a modal)
  • Two-step equation word problem: computers (Opens a modal)
  • Two-step equation word problem: garden (Opens a modal)
  • Two-step equation word problem: oranges (Opens a modal)
  • Two-step equations Get 5 of 7 questions to level up!
  • Two-step equations word problems Get 3 of 4 questions to level up!

Multi-step equations

  • Why we do the same thing to both sides: Variable on both sides (Opens a modal)
  • Intro to equations with variables on both sides (Opens a modal)
  • Equations with variables on both sides: 20-7x=6x-6 (Opens a modal)
  • Equations with parentheses (Opens a modal)
  • Equations with variables on both sides Get 3 of 4 questions to level up!
  • Equations with parentheses Get 3 of 4 questions to level up!

One-step inequalities

  • One-step inequalities examples (Opens a modal)
  • One-step inequalities: -5c ≤ 15 (Opens a modal)
  • One-step inequality word problem (Opens a modal)
  • One-step inequalities review (Opens a modal)
  • One-step inequalities Get 5 of 7 questions to level up!

Two-step inequalities

  • Two-step inequalities (Opens a modal)
  • Two-step inequality word problem: apples (Opens a modal)
  • Two-step inequality word problem: R&B (Opens a modal)
  • Two-step inequalities Get 5 of 7 questions to level up!
  • Two-step inequality word problems Get 3 of 4 questions to level up!

Multi-step inequalities

  • Inequalities with variables on both sides (Opens a modal)
  • Inequalities with variables on both sides (with parentheses) (Opens a modal)
  • Multi-step inequalities (Opens a modal)
  • Multi-step linear inequalities Get 3 of 4 questions to level up!

Solver Title

Practice

Generating PDF...

  • Pre Algebra Order of Operations Factors & Primes Fractions Long Arithmetic Decimals Exponents & Radicals Ratios & Proportions Percent Modulo Number Line Mean, Median & Mode
  • Algebra Equations Inequalities System of Equations System of Inequalities Basic Operations Algebraic Properties Partial Fractions Polynomials Rational Expressions Sequences Power Sums Interval Notation Pi (Product) Notation Induction Logical Sets Word Problems
  • Pre Calculus Equations Inequalities Scientific Calculator Scientific Notation Arithmetics Complex Numbers Polar/Cartesian Simultaneous Equations System of Inequalities Polynomials Rationales Functions Arithmetic & Comp. Coordinate Geometry Plane Geometry Solid Geometry Conic Sections Trigonometry
  • Calculus Derivatives Derivative Applications Limits Integrals Integral Applications Integral Approximation Series ODE Multivariable Calculus Laplace Transform Taylor/Maclaurin Series Fourier Series Fourier Transform
  • Functions Line Equations Functions Arithmetic & Comp. Conic Sections Transformation
  • Linear Algebra Matrices Vectors
  • Trigonometry Identities Proving Identities Trig Equations Trig Inequalities Evaluate Functions Simplify
  • Statistics Mean Geometric Mean Quadratic Mean Average Median Mode Order Minimum Maximum Probability Mid-Range Range Standard Deviation Variance Lower Quartile Upper Quartile Interquartile Range Midhinge Standard Normal Distribution
  • Physics Mechanics
  • Chemistry Chemical Reactions Chemical Properties
  • Finance Simple Interest Compound Interest Present Value Future Value
  • Economics Point of Diminishing Return
  • Conversions Roman Numerals Radical to Exponent Exponent to Radical To Fraction To Decimal To Mixed Number To Improper Fraction Radians to Degrees Degrees to Radians Hexadecimal Scientific Notation Distance Weight Time
  • Pre Algebra
  • One-Step Addition
  • One-Step Subtraction
  • One-Step Multiplication
  • One-Step Division
  • One-Step Decimals
  • Two-Step Integers
  • Two-Step Add/Subtract
  • Two-Step Multiply/Divide
  • Two-Step Fractions
  • Two-Step Decimals
  • Multi-Step Integers
  • Multi-Step with Parentheses
  • Multi-Step Rational
  • Multi-Step Fractions
  • Multi-Step Decimals
  • Solve by Factoring
  • Completing the Square
  • Quadratic Formula
  • Biquadratic
  • Logarithmic
  • Exponential
  • Rational Roots
  • Floor/Ceiling
  • Equation Given Roots
  • Newton Raphson
  • Substitution
  • Elimination
  • Cramer's Rule
  • Gaussian Elimination
  • System of Inequalities
  • Perfect Squares
  • Difference of Squares
  • Difference of Cubes
  • Sum of Cubes
  • Polynomials
  • Distributive Property
  • FOIL method
  • Perfect Cubes
  • Binomial Expansion
  • Negative Rule
  • Product Rule
  • Quotient Rule
  • Expand Power Rule
  • Fraction Exponent
  • Exponent Rules
  • Exponential Form
  • Logarithmic Form
  • Absolute Value
  • Rational Number
  • Powers of i
  • Partial Fractions
  • Is Polynomial
  • Leading Coefficient
  • Leading Term
  • Standard Form
  • Complete the Square
  • Synthetic Division
  • Linear Factors
  • Rationalize Denominator
  • Rationalize Numerator
  • Identify Type
  • Convergence
  • Interval Notation
  • Pi (Product) Notation
  • Boolean Algebra
  • Truth Table
  • Mutual Exclusive
  • Cardinality
  • Caretesian Product
  • Age Problems
  • Distance Problems
  • Cost Problems
  • Investment Problems
  • Number Problems
  • Percent Problems
  • Addition/Subtraction
  • Multiplication/Division
  • Dice Problems
  • Coin Problems
  • Card Problems
  • Pre Calculus
  • Linear Algebra
  • Trigonometry
  • Conversions

Click to reveal more operations

Most Used Actions

Number line.

  • -x+3\gt 2x+1
  • (x+5)(x-5)\gt 0
  • 2x^2-x\gt 0
  • (x+3)^2\le 10x+6
  • \left|3+2x\right|\le 7
  • \frac{\left|3x+2\right|}{\left|x-1\right|}>2
  • What are the 4 inequalities?
  • There are four types of inequalities: greater than, less than, greater than or equal to, and less than or equal to.
  • What is a inequality in math?
  • In math, inequality represents the relative size or order of two values.
  • How do you solve inequalities?
  • To solve inequalities, isolate the variable on one side of the inequality, If you multiply or divide both sides by a negative number, flip the direction of the inequality.
  • What are the 2 rules of inequalities?
  • The two rules of inequalities are: If the same quantity is added to or subtracted from both sides of an inequality, the inequality remains true. If both sides of an inequality are multiplied or divided by the same positive quantity, the inequality remains true. If we multiply or divide both sides of an inequality by the same negative number, we must flip the direction of the inequality to maintain its truth.

inequalities-calculator

  • High School Math Solutions – Inequalities Calculator, Absolute Value Inequalities Part I Last post, we learned how to solve rational inequalities. In this post, we will learn how to solve absolute value... Read More
  • Solve Inequalities Learning Outcomes Describe solutions to inequalities Represent inequalities on a number line Represent inequalities using interval notation Solve single-step inequalities Use the addition and multiplication properties to solve algebraic inequalities and express their solutions graphically and with...
  • 3A.2 Solve Compound Inequalities 3A.2 Learning Objectives Use interval notation to describe intersections and unions Use graphs to describe intersections and unions Solve compound inequalities in the form of or and express the solution graphically and with an interval Express solutions to inequalities graphically and with interval...
  • Linear Inequalities and Systems of Linear Inequalities in Two Variables Learning Objectives Define solutions to a linear inequality in two variables Identify and follow steps for graphing a linear inequality in two variables Identify whether an ordered pair is in the solution set of a linear inequality Define solutions to systems of linear inequalities Graph a system o...
  • Compound Inequalities Learning Objectives Describe sets as intersections or unions Use interval notation to describe intersections and unions Use graphs to describe intersections and unions Solve compound inequalities—OR Solve compound inequalities in the form of or and express the solution graphically and with an inter...
  • Inequalities Introduction to Inequalities Inequalities are used to demonstrate relationships between numbers or expressions. Learning Objectives Explain what inequalities represent and how they are used Key Takeaways Key Points An inequality describes a relationship between two different values. The notation me...
  • Systems of Linear Inequalities 6.3 Learning Objectives Define solutions to systems of linear inequalities Graph a system of linear inequalities and define the solutions region Verify whether a point is a solution to a system of inequalities Identify when a system of inequalities has no solution Solutions from graphs of linear in...
  • 3A.1 Single- and Multi-Step Inequalities 3A.1 Learning Objectives Represent inequalities on a number line Represent inequalities using interval notation Use the addition and multiplication properties to solve algebraic inequalities Express solutions to inequalities graphically, with interval notation, and as an inequality Sometimes there...
  • Reading: Solving One-Step Inequalities Inequalities are similar to equations in that they show a relationship between two expressions. We solve and graph inequalities in a similar way to equations. However, there are some differences that we will talk about in this chapter. The main difference is that for linear inequalities the answer i...
  • Linear Inequalities and Absolute Value Inequalities It is not easy to make the honor role at most top universities. Suppose students were required to carry a course load of at least 12 credit hours and maintain a grade point average of 3.5 or above. How could these honor roll requirements be expressed mathematically? In this section, we will explore...
  • Poverty and Economic Inequality Figure 1. Occupying Wall StreetOn September 17, 2011, Occupy Wall Street began in New York City’s Wall Street financial district. (Credit: modification of work by David Shankbone/Flickr Creative Commons) Occupy Wall Street In September 2011, a group of protesters gathered in Zuccotti Park in New Yo...

Please add a message.

Message received. Thanks for the feedback.

Wyzant

Write and solve inequalities in context

Audrey has $440 to spend at a bicycle store for some new gear and biking outfits. Assume all prices listed include tax. She buys a new bicycle for $259.52. She buys 2 bicycle reflectors for $3.07 each and a pair of bike gloves for $29.12. She plans to spend some or all of the money she has left to buy new biking outfits for $72.61 each. Write and solve an inequality which can be used to determine o, the number of outfits Audrey can purchase while staying within her budget.

3 Answers By Expert Tutors

inequality problem solving question

Kunal P. answered • 3d

Helping Students Shape Their Creativity and Their Minds

No matter what Audrey only has $440 to spend so the total of all her equipment has to be equal to or less than $440.

259.92 + 2(3.07) + 29.12 + 72.61o ≤ 440

since we are trying to figure out how many outfits she can buy at $72.61 for each one thats the number that will be attached to variable o.

Now we add up the total cost so far

72.61o + 294.78 ≤ 440

Now subtract the total cost of items we have so far to see how much remaining money she has left

72.61o ≤ 145.22

Divide by 72.61 to get the number of outfits she can buy

o = 2 outfits

Ash O. answered • 5d

Experienced Mathematics Tutor

We know that Audrey has a total of $440. She can spend all $440 or less than $440, but no more than that. When we set up our inequality, we want to have our equation ≤ 440 since that is the maximum amount of money she can spend.

She buys a new bicycle for $259.52, 2 bicycle reflectors for $3.07 each and a pair of bike gloves for $29.12. That would give us 259.52 + 2 (3.07) + 29.12 so far.

She plans to spend some or all of the money she has left to buy new biking outfits for $72.61 each. We don't know how many outfits she will end up buying, so we will use the variable o to represent the number of outfits she will buy. 72.61 times o will tell us the total cost for all of the outfits she buys.

If we put that all together, we get the following inequality:

259.52 + 2 (3.07) + 29.12 + 72.61 ≤ 440

Mark M. answered • 5d

Math Tutor--High School/College levels

259.52 + 2(3.07) + 29.12 + 72.61o ≤ 440

294.78 + 72.61o ≤ 440

She has exactly enough money left over to purchase 2 outfits.

Still looking for help? Get the right answer, fast.

Get a free answer to a quick problem. Most questions answered within 4 hours.

Choose an expert and meet online. No packages or subscriptions, pay only for the time you need.

RELATED TOPICS

Related questions, how to do i make a graph.

Answers · 4

How do I know my answer?

How do you solve 8x+32/x^2-16.

Answers · 14

12p+15c>360

Answers · 5

Answers · 3

RECOMMENDED TUTORS

inequality problem solving question

find an online tutor

  • Algebra 1 tutors
  • Algebra 2 tutors
  • Algebra tutors
  • College Algebra tutors
  • Hesi tutors
  • CBEST tutors
  • PTCB tutors
  • Precalculus tutors

related lessons

  • Algebra Worksheets List
  • Algebra Study Tips
  • Algebra Calculators
  • Basics of Algebra
  • Algebra Lessons List
  • Preparing for the 2016 SAT Changes
  • Trigonometry / Algebra 2 Trig Help
  • Algebraic and Geometric Patterns

IMAGES

  1. Solving Inequalities (video lessons, examples, solutions)

    inequality problem solving question

  2. Solving inequalities using algebra

    inequality problem solving question

  3. Solving Rational Inequalities Worksheet

    inequality problem solving question

  4. How To Solve Inequalities

    inequality problem solving question

  5. Inequality Algebra Problems with Solutions

    inequality problem solving question

  6. How to Solve Compound Inequalities in 3 Easy Steps

    inequality problem solving question

VIDEO

  1. A Basic Inequality Problem

  2. Linear Inequalities Lec 1

  3. 3 technique to solve inequality 1

  4. Rational Inequality Problem Solving

  5. Mathematics Shorts

  6. discuss the inequality problem || calculus

COMMENTS

  1. Solving Inequality Word Questions

    Solve: Start with: S + A < 9 A = S + 3, so: S + (S + 3) < 9 Simplify: 2S + 3 < 9 Subtract 3 from both sides: 2S < 9 − 3 Simplify: 2S < 6 Divide both sides by 2: S < 3 Sam scored less than 3 goals, which means that Sam could have scored 0, 1 or 2 goals. Alex scored 3 more goals than Sam did, so Alex could have scored 3, 4, or 5 goals. Check:

  2. How to Solve Inequalities—Step-by-Step Examples and Tutorial

    You already know that there are four types of inequalities: > : greater than, < : less than, ≥ : greater than or equal to, and ≤ : less than or equal to (these four types of inequalities are illustrated in Figure 01 above). Basic inequalities do not need to be solved since the variable is already by itself.

  3. Inequalities Practice Questions

    Click here for Answers. inequality. Practice Questions. Previous: Graphical Inequalities Practice Questions. Next: Cumulative Frequency and Box Plot Practice Questions. The Corbettmaths Practice Questions on inequalities.

  4. Solving equations & inequalities

    Quiz Unit test About this unit There are lots of strategies we can use to solve equations. Let's explore some different ways to solve equations and inequalities. We'll also see what it takes for an equation to have no solution, or infinite solutions. Linear equations with variables on both sides Learn

  5. Inequalities word problems (practice)

    Inequalities word problems Google Classroom Kwame must earn more than 16 stars per day to get a prize from the classroom treasure box. Write an inequality that describes S , the number of stars Kwame must earn per day to get a prize from the classroom treasure box. Stuck? Review related articles/videos or use a hint. Report a problem Do 4 problems

  6. Two-step inequalities (practice)

    Two-step inequality word problem: apples. Two-step inequality word problem: R&B. Two-step inequality word problems. Math > 7th grade > Expressions, equations, & inequalities > ... Learn for free about math, art, computer programming, economics, physics, chemistry, biology, medicine, finance, history, and more. Khan Academy is a nonprofit with ...

  7. 1.5: Solve Inequalities

    Substitute the end point 0 into the related equation. Pick a value greater than 0, such as 20, to check in the inequality. (This value will be on the shaded part of the graph.) [/hidden-answer] The previous examples showed you how to solve a one-step inequality with the variable on the left hand side.

  8. Algebra

    For problems 1 - 6 solve each of the following inequalities. Give the solution in both inequality and interval notations. 4(z +2)−1 > 5 −7(4 −z) 4 ( z + 2) − 1 > 5 − 7 ( 4 − z) Solution 1 2(3 +4t) ≤ 6(1 3 − 1 2 t)− 1 4 (2+10t) 1 2 ( 3 + 4 t) ≤ 6 ( 1 3 − 1 2 t) − 1 4 ( 2 + 10 t) Solution −1 <4x+2 <10 − 1 < 4 x + 2 < 10 Solution

  9. PDF Solving Inequalities

    Question 3: The perimeter of the regular pentagon is larger than the perimeter of the equilateral triangle. Form an inequality in terms of x. Solve the inequality to Jind the possible range of values for x. Question 4: Find the range of values of x that satisJies both. Question 5: y is a prime number and also satisJies.

  10. Inequalities

    Solve word problems leading to inequalities of the form px + q > r or px + q < r, where p, q, ... Use this quiz to check your grade 1 to 7 students' understanding of inequalities. 10+ questions with answers covering a range of 1st - 7th grade inequalities topics to identify areas of strength and support! DOWNLOAD FREE. x

  11. Harder linear inequalities & Word problems

    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 − 32t, where the time t is measured in seconds. When will the velocity be between 32 and 64 feet per second (inclusive)?

  12. Algebra

    Here are a set of practice problems for the Solving Equations and Inequalities chapter of the Algebra notes. If you'd like a pdf document containing the solutions the download tab above contains links to pdf's containing the solutions for the full book, chapter and section. At this time, I do not offer pdf's for solutions to individual problems.

  13. Solve inequalities with Step-by-Step Math Problem Solver

    To solve an inequality use the following steps: Step 1 Eliminate fractions by multiplying all terms by the least common denominator of all fractions. Step 2 Simplify by combining like terms on each side of the inequality. Step 3 Add or subtract quantities to obtain the unknown on one side and the numbers on the other.

  14. Linear Inequalities Questions

    Solution: (x - 2)/ (x + 5) > 2 Subtracting 2 from both sides, we get; (x - 2)/ (x + 5) - 2 > 0 [ (x - 2) - 2 (x + 5)]/ (x + 5) > 0 (x - 2 - 2x - 10)/ (x + 5) > 0 - (x + 12)/ (x + 5) > 0 Multiplying -1 on both sides, we get; (x + 12)/ (x + 5) < 0 ⇒ x + 12 < 0 and x + 5 > 0 (or) x + 12 > 0 and x + 5 < 0 ⇒ x < -12 and x > -5 (or) x > -12 and x < -5

  15. Inequalities Practice Questions

    Inequalities Practice Questions section is here now. The following section on Inequalities Practice Questions is complete and fully exhaustive. It covers most of the interesting topics and will let you perfect your problem-solving skills. The following Inequalities Practice Questions section has been divided into various sections for ...

  16. Solving basic equations & inequalities (one variable, linear)

    Algebra (all content) 20 units · 412 skills. Unit 1 Introduction to algebra. Unit 2 Solving basic equations & inequalities (one variable, linear) Unit 3 Linear equations, functions, & graphs. Unit 4 Sequences. Unit 5 System of equations. Unit 6 Two-variable inequalities. Unit 7 Functions. Unit 8 Absolute value equations, functions, & inequalities.

  17. Art of Problem Solving

    Overview. Inequalities are arguably a branch of elementary algebra, and relate slightly to number theory.They deal with relations of variables denoted by four signs: .. For two numbers and : . if is greater than , that is, is positive.; if is smaller than , that is, is negative.; if is greater than or equal to , that is, is nonnegative.; if is less than or equal to , that is, is nonpositive.

  18. Inequality Algebra Problems with Solutions

    Solve the given practice questions based on inequalities. Also, the answer key and explanations are given for the same. Rate Us Views:77208 Instant Access to Free Material Q.1. Solve the inequality 7x+5/3x-5<5. A. (1/3, 0) B. (-∞, 1/3)∪ (5/4, ∞) C. (1/3, 5/4) D. (-∞, 1/3)∪ (5/4, 7) Answer & Explanation Suggested Action

  19. Inequalities on ACT Math: Strategies and Practice

    Typical ACT Inequality Problems. There are three different types of inequality questions you'll see on the ACT, in the order from most to least common: #1: Solve an inequality equation (find the solution set) #2: Identify or answer questions about an inequality graph or number line. #3: Find alternate inequalities that fulfill given information.

  20. 3.7: Solving Systems of Inequalities with Two Variables

    Therefore, to solve these systems we graph the solution sets of the inequalities on the same set of axes and determine where they intersect. This intersection, or overlap, will define the region of common ordered pair solutions. Example 3.7.2: Graph the solution set: {− 2x + y > − 4 3x − 6y ≥ 6.

  21. Linear equations and inequalities

    Unit 2 Algebraic expressions. Unit 3 Linear equations and inequalities. Unit 4 Graphing lines and slope. Unit 5 Systems of equations. Unit 6 Expressions with exponents. Unit 7 Quadratics and polynomials. Unit 8 Equations and geometry. Course challenge. Test your knowledge of the skills in this course.

  22. Inequalities Calculator

    How do you solve inequalities? To solve inequalities, isolate the variable on one side of the inequality, If you multiply or divide both sides by a negative number, flip the direction of the inequality. Show more High School Math Solutions - Inequalities Calculator, Radical Inequalities

  23. Write and solve inequalities in context

    Write and solve an inequality which can be used to determine o, the number of outfits Audrey can purchase while staying within her budget. ... Get a free answer to a quick problem. Most questions answered within 4 hours. OR. Find an Online Tutor Now