Enjoy any 5 free lessons!
You can pick. No account needed.
Watch VideoBecome a member to get full access to our entire library of learning videos, reading material, quiz games, simple DIY activities & more.
Become a member to get full access to our entire library of learning videos, quiz games, & more.
Plans & Pricingto watch this full video.
Access All Videos
and Lessons, No Limits.
Access All Videos
No credit card required,
takes 7 sec to signup.
No card required
Readytogo lessons
that save you time.
Readytogo lessons
If you are on a school computer or network, ask your tech person to whitelist these URLs:
*.wistia.com, fast.wistia.com, fast.wistia.net, embedwistiaa.akamaihd.net
Sometimes a simple refresh solves this issue. If you need further help, contact us.
Find Solutions to Algebraic Inequalities
 Show lesson plan & teacher guide
 Show answers to discussion questions
 Show video only
 Allow visiting of other pages
 Hide assessments
 That an algebraic inequality is a comparison that contains a variable.
 How to solve algebraic inequalities using inverse operations.
 How this knowledge can help us visit a theme park, pack for a trip, and determine how much we earn at a job!

Discussion Questions

Before VideoHow do you solve the equation 5x=30?ANSWER

I divide both sides by 5 to get x=6.

First I subtract 3 from both sides to get [ggfrac]x/2[/ggfrac]=2, and then I multiply both sides by 2 to get x=4. I use inverse operations to solve for the unknown variable.

I can use substitution. Once I find the value of x, I can substitute the value for x in the equation and see if both sides are equal. [ggfrac]4/2[/ggfrac]+3=5, so I know that 2 is the correct value for x.
ANSWER
Less than and greater than. 6 < 8 and 8 > 6.

Students draw a number line with an open point at 5 and the arrow traveling right infinitely. There is not only one possible value for x. Any value that is greater than 5 makes this inequality true. x can be 6, 7, 9.6, 100, or many other values.


After VideoWhich of the following numbers are part of the solution set for x>13? 10, 12, 13, 14, 17ANSWER

Only 14 and 17. x must be greater than 13, so 13 and all smaller numbers are not part of the solution.
ANSWER
15, 16, and 17. Since x must be less than or equal to 17, 17 itself is part of the solution set.

To solve, I treat the > like an equal sign and use inverse operations to solve. Since x is multiplied by 2 and then 3 is added to it, I subtract 3 on both sides and then divide both sides by 2 to get x>2.

Since x is divided by 3 and then 4 is subtracted from it, I first add 4 to both sides, and then multiply both side by 3, to get x≤15.

I can check by substitution. Substituting 6 for x, I get [ggfrac]6/3[/ggfrac]4≤1, which simplifies to 2≤1. Since 2 is less than or equal to 1, I know that 6 satisfies the inequality.



Vocabulary

Equation
DEFINE
Two expressions that have equal value separated by a = sign.

Inequality
DEFINE
A comparison between two amounts using symbols like < and >.

Algebraic inequality
DEFINE
An inequality that has a variable.

Variable
DEFINE
A symbol, like x, used to represent an unknown value.

Inverse operations
DEFINE
Multiplication and division are inverse operations, and addition and subtraction are inverse operations.

Substitution
DEFINE
Plugging in values for x to check if they satisfy an equation or inequality.

Solution of an equation
DEFINE
A value that, when substituted into an equation, yields equal values on both sides of the equal sign.

Solution set of an inequality
DEFINE
A set of values that, when substituted into an inequality, yield a true comparison.

Equation
DEFINE

Reading Material
Download as PDF Download PDF View as Separate PageWHAT ARE ALGEBRAIC INEQUALITIES?Solving algebraic inequalities is similar to solving algebraic equations. You apply inverse operations to isolate the variable on one side. However, there isn’t only one solution to an inequality, but a set of solutions.
To better understand algebraic inequalities…
WHAT ARE ALGEBRAIC INEQUALITIES?. Solving algebraic inequalities is similar to solving algebraic equations. You apply inverse operations to isolate the variable on one side. However, there isn’t only one solution to an inequality, but a set of solutions. To better understand algebraic inequalities…LET’S BREAK IT DOWN!
There are 4 types of inequality symbols.
When talking about inequalities, there are 4 kinds of symbols: < 'less than', > 'greater than', 'less than or equal to', and 'greater than or equal to.' For example, x < 40 means that x can be any value that is less than 40, but it cannot be 40 itself. x 5 means that x can be 5 or any value greater than 5. Try this one yourself: b To enter an amusement park ride, you must be 85 cm or shorter. Represent that as an algebraic inequality, where x is the height of a person who wants to ride.
There are 4 types of inequality symbols. When talking about inequalities, there are 4 kinds of symbols: < 'less than', > 'greater than', 'less than or equal to', and 'greater than or equal to.' For example, x < 40 means that x can be any value that is less than 40, but it cannot be 40 itself. x 5 means that x can be 5 or any value greater than 5. Try this one yourself: b To enter an amusement park ride, you must be 85 cm or shorter. Represent that as an algebraic inequality, where x is the height of a person who wants to ride.Solve an algebraic inequality to pack your suitcase.
You are packing a suitcase for a trip. To go on the plane, your bag needs to weigh less than 100 lbs. You already put some things in your suitcase, and it now weighs 45 lbs. Set up an inequality to represent how much weight can be added. The unknown amount of pounds plus the 45 lbs. already packed must be less than 100 lbs: x+45<100. Solve the inequality using rules that are very similar to solving equations. Treat the less than sign as an equal sign. To isolate the x, subtract 45 from both sides to get x<55. Then the number of pounds you can add to the suitcase must be less than 55! 55 lb or more would be too much. Try this one yourself: Solve the inequality: x – 12 > 30.
Solve an algebraic inequality to pack your suitcase. You are packing a suitcase for a trip. To go on the plane, your bag needs to weigh less than 100 lbs. You already put some things in your suitcase, and it now weighs 45 lbs. Set up an inequality to represent how much weight can be added. The unknown amount of pounds plus the 45 lbs. already packed must be less than 100 lbs: x+45Solve an algebraic inequality to have a successful bake sale.
You made cupcakes to sell at a school bake sale. You spent $48 on ingredients, and you charge $4 for each cupcake. How many cupcakes do you need to sell in order to make more money than you spent? To represent this as an equality, let the number of cupcakes sold be x. The amount of money that you get from customers is 4 times x, and it must be greater than $48 if you want to earn any profit: Solve 4x > 48. Divide both sides by 4 to isolate x: x> 12. Therefore, you need to sell more than 12 cupcakes to make more than you spent. Try this one yourself: Solve the inequality: 3x<42.
Solve an algebraic inequality to have a successful bake sale. You made cupcakes to sell at a school bake sale. You spent $48 on ingredients, and you charge $4 for each cupcake. How many cupcakes do you need to sell in order to make more money than you spent? To represent this as an equality, let the number of cupcakes sold be x. The amount of money that you get from customers is 4 times x, and it must be greater than $48 if you want to earn any profit: Solve 4x > 48. Divide both sides by 4 to isolate x: x> 12. Therefore, you need to sell more than 12 cupcakes to make more than you spent. Try this one yourself: Solve the inequality: 3xSolve algebraic inequalities that are inclusive.
You earn $8 per hour babysitting, plus you already have $12 saved up. How many hours do you need to work to have at least $100 saved up? Represent the amount of money you have with the inequality 8x+12≥100. The words "at least" say that 100 is included. Solve for x using inverse operations. First, subtract 12 from both sides to get 8x≥88. Next, divide both sides by 8 to get x≥11. That means that you need to work 11 hours or more to have at least $100. 11 is included in the answer, since working 11 hours means you have saved $100. Try this one yourself: Solve the inequality: 3x+9≥63.
Solve algebraic inequalities that are inclusive. You earn $8 per hour babysitting, plus you already have $12 saved up. How many hours do you need to work to have at least $100 saved up? Represent the amount of money you have with the inequality 8x+12≥100. The words "at least" say that 100 is included. Solve for x using inverse operations. First, subtract 12 from both sides to get 8x≥88. Next, divide both sides by 8 to get x≥11. That means that you need to work 11 hours or more to have at least $100. 11 is included in the answer, since working 11 hours means you have saved $100. Try this one yourself: Solve the inequality: 3x+9≥63.Solve algebraic inequalities with fractions.
Your band is playing a concert, and you get to keep [ggfrac]1/3[/ggfrac] of the money made from ticket sales. You promised a friend $10 for helping move your gear. How much money needs to be made from ticket sales for you to have $100 at the end of the night? Let x represent the total earned from ticket sales. Then we can represent your earnings by the inequality [ggfrac]x/3[/ggfrac]10≥100. To solve, add 10 to both sides then multiply both sides by 3. [ggfrac]x/3[/ggfrac]≥110. x=330. This means that you need to sell at least $330 worth of tickets to make at least $100. Try this one yourself: Solve the inequality x2+13≥27.
Solve algebraic inequalities with fractions. Your band is playing a concert, and you get to keep [ggfrac]1/3[/ggfrac] of the money made from ticket sales. You promised a friend $10 for helping move your gear. How much money needs to be made from ticket sales for you to have $100 at the end of the night? Let x represent the total earned from ticket sales. Then we can represent your earnings by the inequality [ggfrac]x/3[/ggfrac]10≥100. To solve, add 10 to both sides then multiply both sides by 3. [ggfrac]x/3[/ggfrac]≥110. x=330. This means that you need to sell at least $330 worth of tickets to make at least $100. Try this one yourself: Solve the inequality x2+13≥27. 
Practice Word Problems

Practice Number Problems

Teacher Resources
These downloadable teacher resources can help you create a full lesson around the video. These PDFs incorporate using class discussion questions, vocabulary lists, printable math worksheets, quizzes, games, and more.
Select a Google Form
Choose a way to play this quiz game

Questions appear on the teacher's screen. Students answer on their own devices.
Start a Free Trial Today. Get a $5 Amazon Gift Card!
Teachers! Start a free trial & we'll send your gift card within 1 day. Only cards left. Try it now.
This email is associated with a Science Kit subscription. Kit subscriptions are managed on this separate page: Manage Subscription

Science & Math$_{/yr}

Science Only$_{/yr}
Solve x3>24.
Solve 3x+5≤14.
Solve [ggfrac]x/5[/ggfrac]+17≥19.
access all lessons
• No credit card required •
"My students loved the videos. I started the video subscription in May and used them as a review before the state test, which I know contributed to 100% of my class passing the state test."
Rhonda Fox 4th Grade Teacher, Ocala, Florida• No credit card required •
"My students loved the videos. I started the video subscription in May and used them as a review before the state test, which I know contributed to 100% of my class passing the state test."
Rhonda Fox 4th Grade Teacher, Ocala, Florida• No credit card required •
Already a member? Sign In
* no credit card required *
* no credit card required *
* no credit card required *
no credit card required
Skip, I will use a 3 day free trial
Enjoy your free 30 days trial

Unlimited access to our full library
of videos & lessons for grades K5. 
You won’t be billed unless you keep your
account open past your 14day free trial. 
You can cancel anytime in 1 click on the
manage account page or by emailing us.

Unlimited access to our full library of videos & lessons for grades K5.

You won't be billed unless you keep your account open past 14 days.

You can cancel anytime in 1click on the manage account page.
Cancel anytime in 1click on the manage account page before the trial ends and you won't be charged.
Otherwise you will pay just $10 CAD/month for the service as long as your account is open.
Cancel anytime on the manage account page in 1click and you won't be charged.
Otherwise you will pay $10 CAD/month for the service as long as your account is open.
We just sent you a confirmation email. Enjoy!
DonePlease login or join.