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.

## College Algebra

### Unit 1: Lesson 4

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"

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
A number line from negative two to eight by ones. There is an open circle at three with an arrow to the left. There is an open circle at five with an arrow to the right.
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:
A number line from negative two to eight by ones. There is an open circle at zero with an arrow to the right. There is an open circle at four with an arrow to the left.
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:
A number line from negative one to five by ones. There is an open circle at one with an arrow to the right. There is a closed circle at two with an arrow to the right.
So our compound inequality can be expressed as the simple inequality:
x, is greater than, 1

### 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:
A number line from negative three to three by one. There is an open circle at negative one with an arrow to the left and to the right.
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.

## Practice

Problem 1
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

Want more practice? Check out this exercise.

## Want to join the conversation?

• 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.
• 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!
• 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.

I hope this helps, please tell me if you have any more questions and please ask more questions.

😏😊😉😄😃😅🙄🙄🙄

• 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.