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First things first, let's remember what a compound inequality is. A compound inequality is two separate inequalities that are joined together by "and" or "or."

If you're more of an English/grammar person, think of a compound inequality like a compound sentence. There are two independent clauses joined together by words like "and" and "or." It's the same thing in math, except replace the clauses with inequalities.

Let's look at two examples of how to solve word problems with compound inequalities. We'll look at one example with "and" and one example with "or."

Example 1:

There are a few things to notice about this word problem before we set up our compound inequality.

The phrase "between $10$ and $50$ percent" tells us that this is going to be a compound inequality with the connector "and." We'll look at how to write that in a minute.

The other thing to notice is the word inclusive. This means that $10$ and $50$ (the two ends of the range) need to be included as possible solutions themselves. To show that in our inequality, we just need to use $≤$ and $≥$ (instead of $<$ and $>$).

The first thing we need to do is figure out what the end points are for what the ant can carry. We know its bodyweight is $5\text{ mg}$, so we need to find out what $10\%$ and $50\%$ of that value are.

$10$ of $5$ is $5 \cdot\frac{10}{100}$, which is $0.5$.

$50$ of $5$ is $5 \cdot\frac{50}{100}$, which is $2.5$.

We know that the ant can carry between $0.5\text{ mg}$ and $2.5\text{ mg}$, inclusive. We need to write an inequality that represents that information.

An ant can carry an amount ($x$) if it's between $0.5$mg and $2.5$mg, inclusive, so our inequality is:

$$0.5 \le x \le 2.5$$

If we wanted to write this in interval notation, it would look like this: $[0.5, 2.5]$. Notice the brackets that we used to indicate "inclusive."

We can also verify that this inequality makes sense by plotting our range on a number line.

Example 2:

First, let's review the problem for any signal words. One thing to note is that this problem uses the phrase "colder than $45$ degrees or warmer than $80$ degrees." This means this will be an "or"-style compound inequality.

Also notice that the problem is asking us for the answer in Celsius, not Fahrenheit.

Last, there's no mention of "inclusive." This means that we will be using $<$ and $>$ signs.

Let's start by converting the two endpoints ($45$ and $80$) from Fahrenheit to Celsius. Here's the formula to convert between Fahrenheit and Celsius:

$$^{\circ}C = \frac{^{\circ}F - 32}{1.8}$$

$$\begin{align} ^{\circ}C &= \frac{45 - 32}{1.8}\\ ^{\circ}C &= \frac{13}{1.8}\\ ^{\circ}C &= 7.2\\ \end{align}$$

$$\begin{align} ^{\circ}C &= \frac{80 - 32}{1.8}\\ ^{\circ}C &= \frac{48}{1.8}\\ ^{\circ}C &= 26.7\\ \end{align}$$

Now we can write two separate inequalities to show the possible temperature of Jim's pool water.

To write the inequality that describes the temperature of Jim's water, we would use both those equations and say:

$$x < 7.2\ or\ x > 26.7$$

This tells us that Jim's water is either less than $7.2$ degrees Celsius or greater than $26.7$ degrees Celsius.

The interval notation for this solution looks like this: $(-\infty, 7.2)\text{ or }(26.7, \infty)$.

This can be also written as $(-\infty, 7.2) ∪ (26.7, \infty)$. The $∪$ symbol indicates the union of sets—which is a fancy way of saying that the solution can be in either of the two ranges we wrote.

We can also verify this using a number line.

Breaking Down Compound Inequality Word Problems

Word problems can be tricky, but we can conquer them! These kinds of problems follow patterns that can help find the answer.

Inclusivity

A good first step to take is to find out if the final answer will be inclusive (using signs such as $\le, \ge, [ \phantom{0} ]$) or exclusive (using signs such as $<, >, ()$ ). Some problems may actually say the words inclusive or exclusive, but not all of them. Statements like "at least" or "at most" can imply that the answer is inclusive. Sometimes you may need to use the context of the problem to decide if it is inclusive or exclusive.

Let's try one problem to practice finding exclusivity.

AND vs OR

The next step is to figure out what kind of interval you're looking for. If you see and, you'll be looking for an interval between two numbers. If you see or, you'll be looking for a range below one number, and above a different one. You may need to use context clues to figure out which one is implied. Many times the words will be right in the problem.

Once you've figured out what kind of answer you are trying to find, solving the problem is the same as solving any other kind of inequality.

(Video) Compound Inequalities in Word Problems

by Kara K McWilliams

This video by Kara K McWilliams works through some compound inequality word problems.

$\large{\text{Summary}}$

Notice how she figures out when to use $<, >$ or $\leq$, $\geq$.

Inclusive is a word that pretty much always means that you'll need to use the "or equal to" symbols ($\le$ or $\ge$). "Inclusive" won't always be there when you should use those symbols, but it's a pretty good indication.

If you see and in a problem, it probably means that $x$ will be in a range between two numbers.

If you see or in a problem, it probably means that $x$ will be in two ranges.

Example

Kara works through the following problem:

Jill wants to find her test average in her algebra class. The scores that she made on her tests were $75, 82, 78,$ and $85$. What does Jill need to score on her next test if she wants her average to be between $78$ and $80,$ non-inclusive?

Jill wants her score to be higher than $78$ and lower than $80$. Let's set this up as a compound inequality

$$78<\text{average}<80$$

We are going to plug all of Jill's grades into a formula to find the average.

$$\begin{align}\text{average} &= \frac{75+82+78+85+x}{5}\\ \text{average}&=\frac{320+x}{5}\end{align}$$

Let's plug this into the inequality.

$$78<\frac{320+x}{5}<80$$

We'll solve each part of the inequality seperately.

$$\begin{align} 78&<\frac{320+x}{5}\\ 390&<320+x\\ 70&<x\\ \\ \frac{320+x}{5}&<80\\ 320+x&<400\\ x&<80\end{align}$$

Jill's test score needs to be above a $70$ and below an $80$ to end with an average between $78$ and $80$. Here's our answer written in both inequality form and interval notation:

$$70<x<80\\ (70,80)$$