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## Algebra 1

### Course: Algebra 1 > Unit 2

Lesson 6: Compound inequalities# Compound inequalities review

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"

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"

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:

## Solving compound inequalities

### Example with "OR"

**Solve for x.**

Solving the first inequality for x, we get:

Solving the second inequality for x, we get:

Graphically, we get:

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

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

### Example with "AND"

**Solve for x.**

Solving the first inequality for x, we get:

Solving the second inequality for x, we get:

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.*

## Want to join the conversation?

- how do you solve it with fracttions involved?(6 votes)
- 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.(35 votes)

- 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.(22 votes)
- So in a OR inequality, the answer can not be “there are no solutions?(11 votes)
- 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.(15 votes)

- 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?(10 votes)
- 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.(5 votes)

- 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!(10 votes)
- 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.(7 votes)- 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.(11 votes)

- Does the presence of a solution to an inequality depend on whether if the inequality is an 'OR' and an 'AND'?(3 votes)
- 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)(14 votes)

- 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.(5 votes)
- 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.(2 votes)

- How would I write an "or" compound inequality?(5 votes)
- You just write OR between the inequality like

4>x OR x>2 It is not like and were it is 4>x>2(4 votes)

- 8v - 7 ≤ 10 - 9v ≤ -8v + 10(4 votes)
- 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.(6 votes)