So in a OR inequality, the answer can not be “there are no solutions?

That's correct. Any solution from either individual inequality becomes a solution for the compound inequality. Similarly, with an AND inequality, you can't get a solution of all real numbers because the AND restricts the solution set to only those value in common to both individual inequalities.

I am really struggling trying to solve these kinds of problems. When solving OR or AND, what exactly is the solution for the question? Because every time I solve one of these problems I get it wrong but there is no explanation for why. Can someone give me an answer to why?

sometimes I was confused at which way the sign would or is suppose to go. I've noticed that if the bottom of the fraction is minus, then the sign flips. However, if its just the top, it doesn't flip. As long as their is a negative number on the bottom, it seems to always flip the equality sign. Also, if both are negative (top and bottom) then the fraction turns positive but because the negative sign is on the bottom the sign will still flip. OR is more for either or. But AND, has to include a number that will overlap. If it doesn't overlap then it is NO SOLUTIONS and if the number cancels itself out I.E. Greater then one, less than one, then there are also no solutions. I have to literally write the line graph out every time to get it right.

I don't quite understand the union part of compound inequalities. Even if I find the union of an AND inequality, it seems like more often than not the union solution does not apply to both of the inequalities.

compound inequalities are algebra inequalities but, use a line plot. So a inequality first should be solved. Then you must make a line plot. Then you must plot the line plot and show that it is greater than, less than and something like that. Anyway after the line plot is done you must answer the question meaning what x or z or y or ect. means. I do not know what the *union* you are talking about. Please explain.* I hope this helps, please tell me if you have any more questions and please ask more questions. 😏😊😉😄😃😅🙄🙄🙄 Ask more questions.

how do you solve it with fracttions involved?

I'll take an example here. Let's say you have 3 <1/2x + 5 < 17. So you would solve this by first subtracting five from both sides using the Subtraction Property of Equality, which would look like this: 3 - 5 < 1/2x + 5 - 5 < 17 - 5 --> -2 < 1/2x < 12. From there, you want to change the 1/2x into x (or 1x), and to do that, you would multiply the fraction to get 1 in order to isolate the variable (for instance, 2/3 would be multiplied by 3/2 to get 1), so the end result would look like this: -2*2 < 1/2x*2 < 12*2, which would lead to the answer of -4 < x < 24.

Can someone give an example of AND that has "many solutions"? I'll be the one to graph it. I just need equations. Thank you.

It could also be as simple as -6 ≤ 2x + 2 ≤ 8 which says that 2x + 2 ≥ - 6 AND 2x + 2 ≤ 8. To solve, subtract 2 all the way across, -8 ≤ 2x ≤ 6, then divide by 2 to get -4 ≤ x ≤ 3. Infinite amount of numbers between -4 and 3.

How would I write an "or" compound inequality?

You just write OR between the inequality like 4>x OR x>2 It is not like and were it is 4>x>2

8v - 7 ≤ 10 - 9v ≤ -8v + 10

Break it into two parts and put back together later. So 8v-7≤10-9y and 10-9v≤-8v+10. On the first, add 9y and add 7 to get 17v ≤ 17, so v ≤1. On the second, you add 9v and subtract 10 to get 0≤v. Putting these together as a union, 0≤v≤1. Values in the range such as 0 gives -7≤10≤10 which is a true statement and 1 gives 1≤1≤2 which is also true. Trying -1 (-15≤19≤18) and 2 (9≤-8≤-6) both of which are false.

Does the presence of a solution to an inequality depend on whether if the inequality is an 'OR' and an 'AND'?

Yes! The AND and OR are operations themselves. The AND tells you to find the intersection of the 2 solution sets (final solution is only the solutions in common to both inequalities). The OR tells you to find the union of the 2 solutions sets (all possible solutions to either inequality)

Main content

Algebra 1

Course: Algebra 1 > Unit 2

Lesson 6: Compound inequalities

Compound inequalities review

Q: Can someone give an example of AND that has "many solutions"? I'll be the one to graph it. I just need equations. Thank you.

It could also be as simple as -6 ≤ 2x + 2 ≤ 8 which says that 2x + 2 ≥ - 6 AND 2x + 2 ≤ 8. To solve, subtract 2 all the way across, -8 ≤ 2x ≤ 6, then divide by 2 to get -4 ≤ x ≤ 3. Infinite amount of numbers between -4 and 3.

Q: How would I write an "or" compound inequality?

You just write OR between the inequality like 4>x OR x>2 It is not like and were it is 4>x>2

Google Classroom

A compound inequality is an inequality that combines two simple inequalities. This article provides a review of how to graph and solve compound inequalities.

What is a compound inequality?

A compound inequality is an inequality that combines two simple inequalities. Let's take a look at some examples.

Example with "OR"

x, is less than, 3, space, start color #7854ab, start text, space, O, R, space, end text, end color #7854ab, space, x, is greater than, 5

So, for example, the numbers 0 and 6 are both solutions of the compound inequality, but the number 4 is not a solution.

Example with "AND"

x, is greater than, 0, space, start color #e07d10, start text, space, A, N, D, space, end text, end color #e07d10, space, x, is less than, 4

This compound inequality is true for values that are both greater than zero and less than four. Graphically, we represent it like this:

So, in this case, 2 is a solution of the compound inequality, but 5 is not because it only satisfies one of the inequalities, not both.

Note: If we wanted to, we could write this compound inequality more simply like this:

0, is less than, x, is less than, 4

Solving compound inequalities

Example with "OR"

Solve for x.

2, x, plus, 3, is greater than or equal to, 7, space, start color #7854ab, start text, space, O, R, space, end text, end color #7854ab, space, 2, x, plus, 9, is greater than, 11

Solving the first inequality for x, we get:

\begin{aligned} 2x+3 &\geq 7 \\\\ 2x &\geq 4 \\\\ x &\geq 2 \end{aligned}

Solving the second inequality for x, we get:

\begin{aligned} 2x+9&>11 \\\\ 2x&>2\\\\ x&>1 \end{aligned}

Graphically, we get:

So our compound inequality can be expressed as the simple inequality:

x, is greater than, 1

Want to learn more about compound inequalities that use OR statements? Check out this video.

Example with "AND"

Solve for x.

4, x, minus, 39, is greater than, minus, 43, space, start color #e07d10, start text, space, A, N, D, end text, end color #e07d10, space, 8, x, plus, 31, is less than, 23

Solving the first inequality for x, we get:

\begin{aligned}4x-39&> -43 \\\\ 4x &> -4 \\\\ x &>-1 \end{aligned}

Solving the second inequality for x, we get:

\begin{aligned} 8x+31&<23\\\\ 8x&<-8\\\\ x&<-1 \end{aligned}

Graphically, we get:

Strangely, this means that there are no solutions to the compound inequality because there's no value of x that's both greater than negative one and less than negative one.

Want to learn more about compound inequalities that use AND statements? Check out this video.

Practice

Problem 1

Current

Solve for x.

5, x, minus, 4, is greater than or equal to, 12, space, start text, space, O, R, space, end text, space, 12, x, plus, 5, is less than or equal to, minus, 4

(Choice A)
x, is greater than or equal to, start fraction, 16, divided by, 5, end fraction or x, is less than or equal to, minus, start fraction, 3, divided by, 4, end fraction
(Choice B)
x, is less than or equal to, start fraction, 16, divided by, 5, end fraction
(Choice C)
x, is greater than or equal to, minus, start fraction, 3, divided by, 4, end fraction
(Choice D)
There are no solutions
(Choice E)
All values of x are solutions

Want more practice? Check out this exercise.

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Miguel Monteon :/
Posted 6 years ago. Direct link to Miguel Monteon :/'s post “how do you solve it with ...”
how do you solve it with fracttions involved?
Button navigates to signup pageComment on Miguel Monteon :/'s post “how do you solve it with ...”
(6 votes)
Answer
- Agent Raf
  Posted 6 years ago. Direct link to Agent Raf's post “I'll take an example here...”
  I'll take an example here. Let's say you have 3 <1/2x + 5 < 17. So you would solve this by first subtracting five from both sides using the Subtraction Property of Equality, which would look like this: 3 - 5 < 1/2x + 5 - 5 < 17 - 5 --> -2 < 1/2x < 12. From there, you want to change the 1/2x into x (or 1x), and to do that, you would multiply the fraction to get 1 in order to isolate the variable (for instance, 2/3 would be multiplied by 3/2 to get 1), so the end result would look like this: -2*2 < 1/2x*2 < 12*2, which would lead to the answer of -4 < x < 24.
  Button navigates to signup page
  (35 votes)
Owen Choi
Posted 4 months ago. Direct link to Owen Choi's post “Instead of focusing on wh...”
Instead of focusing on why math is bad, let's explore how we can improve our understanding and appreciation of this important subject. Math is an essential part of our daily lives, and it helps us understand the world around us. It is a valuable tool for problem-solving, critical thinking, and decision-making. Is there anything specific about math that you find challenging or confusing? I'm here to help and provide guidance.
Button navigates to signup pageComment on Owen Choi's post “Instead of focusing on wh...”
(22 votes)
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person
Posted 5 months ago. Direct link to person's post “So in a OR inequality, th...”
So in a OR inequality, the answer can not be “there are no solutions?
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(11 votes)
Answer
- Kim Seidel
  Posted 5 months ago. Direct link to Kim Seidel's post “That's correct. Any solu...”
  That's correct. Any solution from either individual inequality becomes a solution for the compound inequality.
  
  Similarly, with an AND inequality, you can't get a solution of all real numbers because the AND restricts the solution set to only those value in common to both individual inequalities.
  Button navigates to signup page
  (15 votes)
Halie Tarbell
Posted 3 months ago. Direct link to Halie Tarbell's post “I am really struggling tr...”
I am really struggling trying to solve these kinds of problems. When solving OR or AND, what exactly is the solution for the question? Because every time I solve one of these problems I get it wrong but there is no explanation for why. Can someone give me an answer to why?
Button navigates to signup pageComment on Halie Tarbell's post “I am really struggling tr...”
(10 votes)
Answer
- April Turgutalp
  Posted 3 months ago. Direct link to April Turgutalp's post “sometimes I was confused ...”
  sometimes I was confused at which way the sign would or is suppose to go. I've noticed that if the bottom of the fraction is minus, then the sign flips. However, if its just the top, it doesn't flip. As long as their is a negative number on the bottom, it seems to always flip the equality sign. Also, if both are negative (top and bottom) then the fraction turns positive but because the negative sign is on the bottom the sign will still flip. OR is more for either or. But AND, has to include a number that will overlap. If it doesn't overlap then it is NO SOLUTIONS and if the number cancels itself out I.E. Greater then one, less than one, then there are also no solutions. I have to literally write the line graph out every time to get it right.
  Comment on April Turgutalp's post “sometimes I was confused ...”
  (5 votes)
Bart Kornacki
Posted 6 years ago. Direct link to Bart Kornacki's post “Why didn't you swap an in...”
Why didn't you swap an inequality with a sign in problem 2 practice while dividing by negative number (-7). This has changed the equation!
Button navigates to signup pageComment on Bart Kornacki's post “Why didn't you swap an in...”
(10 votes)
Answer
nekk
Posted 5 years ago. Direct link to nekk's post “I don't quite understand ...”
I don't quite understand the union part of compound inequalities.
Even if I find the union of an AND inequality, it seems like more often than not the union solution does not apply to both of the inequalities.
Button navigates to signup pageButton navigates to signup page
(7 votes)
Answer
- STiger24
  Posted 5 years ago. Direct link to STiger24's post “compound inequalities are...”
  compound inequalities are algebra inequalities but, use a line plot.
  
  So a inequality first should be solved.
  
  Then you must make a line plot.
  
  Then you must plot the line plot and show that it is greater than, less than and something like that.
  
  Anyway after the line plot is done you must answer the question meaning what x or z or y or ect. means.
  
  I do not know what the union you are talking about.
  
  Please explain.*
  
  I hope this helps, please tell me if you have any more questions and please ask more questions.
  
  😏😊😉😄😃😅🙄🙄🙄
  
  Ask more questions.
  Comment on STiger24's post “compound inequalities are...”
  (11 votes)
yoonjuliagil
Posted 3 years ago. Direct link to yoonjuliagil's post “Does the presence of a so...”
Does the presence of a solution to an inequality depend on whether if the inequality is an 'OR' and an 'AND'?
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(3 votes)
Answer
- Kim Seidel
  Posted 3 years ago. Direct link to Kim Seidel's post “Yes! The AND and OR are ...”
  Yes! The AND and OR are operations themselves. The AND tells you to find the intersection of the 2 solution sets (final solution is only the solutions in common to both inequalities). The OR tells you to find the union of the 2 solutions sets (all possible solutions to either inequality)
  Comment on Kim Seidel's post “Yes! The AND and OR are ...”
  (14 votes)
Josh
Posted 6 years ago. Direct link to Josh's post “Can someone give an examp...”
Can someone give an example of AND that has "many solutions"? I'll be the one to graph it. I just need equations. Thank you.
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(5 votes)
Answer
- David Severin
  Posted 6 years ago. Direct link to David Severin's post “It could also be as simpl...”
  It could also be as simple as -6 ≤ 2x + 2 ≤ 8 which says that 2x + 2 ≥ - 6 AND 2x + 2 ≤ 8. To solve, subtract 2 all the way across, -8 ≤ 2x ≤ 6, then divide by 2 to get -4 ≤ x ≤ 3. Infinite amount of numbers between -4 and 3.
  Comment on David Severin's post “It could also be as simpl...”
  (2 votes)
Christy Brokaw
Posted 5 years ago. Direct link to Christy Brokaw's post “How would I write an "or"...”
How would I write an "or" compound inequality?
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(5 votes)
Answer
- Ben
  Posted 5 years ago. Direct link to Ben's post “You just write OR between...”
  You just write OR between the inequality like
  
  4>x OR x>2 It is not like and were it is 4>x>2
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  (4 votes)
Ethan hernandez
Posted 4 years ago. Direct link to Ethan hernandez's post “8v - 7 ≤ 10 - 9v ≤ -8v + ...”
8v - 7 ≤ 10 - 9v ≤ -8v + 10
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(4 votes)
Answer
- David Severin
  Posted 4 years ago. Direct link to David Severin's post “Break it into two parts a...”
  Break it into two parts and put back together later. So 8v-7≤10-9y and 10-9v≤-8v+10. On the first, add 9y and add 7 to get 17v ≤ 17, so v ≤1. On the second, you add 9v and subtract 10 to get 0≤v. Putting these together as a union, 0≤v≤1. Values in the range such as 0 gives -7≤10≤10 which is a true statement and 1 gives 1≤1≤2 which is also true. Trying -1 (-15≤19≤18) and 2 (9≤-8≤-6) both of which are false.
  Button navigates to signup page
  (6 votes)