Solving Compound Inequalities - Examples & Practice

Oh no! You missed all of your favorite college basketball games last weekend because your mom had you too busy doing chores to watch any TV! You stink mom!

After you pout in anger at the situation, your dad gives you some clues about the score of the game you wanted to see. In order to understand his clues, we'll need to understand how to combine inequalities using the words "AND," and "OR".

You were eager to watch University of Minnesota play against the University of Indiana in men's basketball on Saturday, February 16, 2019. Your dad leaves clues for you to guess how many points each team scored.

He says, "let $I$ be the score of the University of Indiana." I'll tell you that Indiana scored more than $62$ points, and if you double their score, it wasn't more than $126$ points.

The goal is to find the value of $I$. This is known as a compound inequality with "AND," because the description contains the word "AND." We'll turn this into a compound inequality and then solve it. We call these words connecting words.

We are given two pieces of information that both must be true. Since Indiana scored more than $62$ points, this is a STRICT inequality. We know that $I > 62.$ At the same time, we know that double the score is no more than $126.$ This inequality, unlike the previous one, is non-strict, so we represent it as $2I \leq 126.$ In summary, we have two pieces of information.

We write a compound inequality as the two inequalities with a connecting word in the middle:

$$I > 62\text{ AND }2I \leq 126$$

What do we do? We solve all of the inequalities to get information about $I$. The first inequality is already simplified, but we can do a bit of work on the second inequality to rewrite it.

For now, just forget about the first inequality for a bit, and we'll solve

$$2I \leq 126$$

As you might remember, our goal is to isolate the variable. We do this by dividing both sides of the inequality by $2$ to get

$$I \leq 63.$$

We have found a new way to write $2I \leq 126,$ so we replace it in the compound inequality:

$$I > 62\text{ AND }I \leq 63$$

What numbers satisfy both inequalities? We need a number more than $62$, and a number not more than $63$. Also, since these are basketball scores we're talking about, we'd surely bet the scores were integers. The only integer satisfying both inequalities is $63$, so we conclude Indiana scored $63$ points.

"Not bad," your Dad chuckles. "You seem to remember what you learned in math class!"

You smile back. Knowing your dad, it will only get harder. He hasn't yet given you any clues about the University of Minnesota's score.

"Ok, so let me help you out in figuring out the University of Minnesota's score in this game. Either six points more than their score was less than $57$ points, or half their score was at most $42$ points. Let $M$ be Minnesota's score. Can you tell me what $M$ is?"

You accept the challenge and begin moving towards solving another riddle.

Let's set up another compound inequality. The strategy is the same as before:

• Identify each regular inequality that comprises the compound inequality.

• Decide whether to use "AND" or "OR."

The first piece of information is "six points more than Minnesota's score was less than $57$ points". The word "more" implies addition, and "less than" implies use of the strict inequality $<$. Piecing these elements together gives us

$$M + 6 < 57$$

What about the second inequality? Dad gave it to us in words: "half their score was at most $42$ points." Here, the word "half" implies division, and "at most" implies the use of the non-strict inequality $\leq$. So our second regular inequality is

$$\frac{M}{2} \leq 42.$$

How do we construct the compound inequality? We need to decide which connecting word to use: "AND" or "OR."

We connect the two regular inequalities with the connecting word:

$$M + 6 < 57\text{ OR }\frac{M}{2} \leq 42$$

We need to solve each inequality individually to get information about $M$. Let's try the first one.

$$M + 6 < 57$$

On the same side of the inequality as $M$ is an addition by $6$. We isolate the variable $M$ by undoing this. In other words subtracting $6$ from both sides of the inequality.

$$M + 6 - 6 < 57 - 6$$

Simplifying gives

$$M < 51$$

Now we replace $M + 6 < 57$ with $M < 51$ in the compound inequality.

$$M < 51\text{ OR }\frac{M}{2} \leq 42$$

Let's address the other regular inequality.

$$\frac{M}{2} \leq 42$$

This time, the operation needed to isolate $M$ in the inequality is multiplication by $2$. Doing so gives

$$M \leq 84$$

In summary, our compound inequality becomes

$$M < 51\text{ OR }M \leq 84$$

There is a problem. This is all correct, but it isn't enough information to determine Minnesota's score in the game. In general,

$$\color{#8925AE}{\text{Compound inequalities don't always tell you what your variable equals!}}$$

Before your Dad can praise you for deciphering his riddles, you storm back to your room and pout once again. He sits and laughs, greatly satisfied with himself. He knew you thought you'd find what $M$ is after solving his compound inequality.

Epilogue: Your birthday comes around two months later, for which your parents decide to get you your first cell phone as a present.

The first thing you search for is the score of the Indiana-Minnesota basketball game, and find this website.

It turns out Minnesota had won the basketball game $84-63$! Fittingly enough, your parents buy you a University of Minnesota t-shirt as a second present.

(Video) Solving Compound Inequalities

by Math Meeting

This video by Math Meeting outlines the steps to solve compound inequalities.

Summary

The first example works through

$$2x-5<-11 \text{ or } -3x+14\leq 2$$

The first step for these kinds of inequalities is to solve each of them separately for $x$. We can start with the left side.

$$\begin{align*} 2x-5&<-11 \text{ } &&\\ 2x&<-6 \text{ } && \text{adding 5 to both sides}\\ x&<-3 \text{ } && \text{dividing both sides by 2} \end{align*}$$

Now, for the right side:

$$\begin{align*} -3x+14 &\leq 2 \text{ } &&\\ -3x&\leq -12 \text{ } && \text{subtracting 14 from both sides}\\ x&\geq 4 \text{ } && \text{dividing both sides by -3} \end{align*}$$

The compound inequality is equivalent to $x<-3$ or $x\geq 4$. In interval notation, this corresponds to $(-\infty, -3) \cup [4, \infty)$.

The second inequality is $-5 < 3x-8 \leq 1$. There is no "and" in this statement. However, this inequality is the same as saying $-5 <3x-8$ and $3x-8 \leq 1$. We can solve this in the first form as follows:

$$\begin{align*} -5 &< 3x-8 \leq 1 \text{ } &&\\ 3 &< 3x \leq 9 \text{ } &&\text{adding 8 to all three parts}\\ 1 &< x \leq 3 \text{ } &&\text{diving all three parts by 3}\\ \end{align*}$$

In interval notation, this is the same as $(1,3]$.

Compound Inequalities

We have learned the methods of manipulating and solving inequalities. We have also learned about compound inequalities and their word problems.

The key to solving word problems is performing two important steps: interpreting scenarios correctly, and then translating them into inequalities.

There are two kinds of compound inequalities: intersection (AND) & union (OR).

Intersection (AND)

Let's break down AND inequalities.

"$x$ is less than or equal to $2$ and $x$ is greater than or equal to $-4$."

This is the same as saying:

$x$ has to be less than or equal to $2$. $x$ also has to be greater than or equal to $-4$.

Imagine a number line and try drawing out the two regions: $\\$

1. less than or equal to $2$ $\\$
2. greater than or equal to $-4$

We see there is an overlap of the two regions. $x$ has to satisfy two conditions, it is at most $2$ and at least $-4$. A valid $x$-value is in the intersection of the two regions.

Union (OR)

Let's break down OR inequalities.

Personally, when I hear "union", I intuitively connect it to "and". That is not quite the case in math. For example:

"$x$ is less than or equal to $-2$ or $x$ is greater than or equal to $1$."

This is the same as saying:

"$x$ COULD be less than or equal to $-2$. $x$ also COULD be greater than or equal to $1$."

Imagine a number line and try drawing out the two regions: $\\$

1. less than or equal to $-2$ $\\$
2. greater than or equal to $1$

There is no overlap between the two regions. Rather, $x$ could be at most $-2$ or at least $1$. A valid $x$ value is in the union, or combination, of these two regions.

Another way to think about this is that values greater than $-2$ (exclusive) and less than $1$ (exclusive) are not a part of the solution. All other real numbers are part of the solution.

Notice that the numbers and signs also play a role in defining whether a compound inequality refers to a union or an intersection. When in doubt, draw things out on a number line, and go from there!

We know that if one of the items is more than $\$4$, it will sell very slowly. So, we want:

$$x\le 4$$

We also know that the profit is not satisfactory if the price is below $\$2.50$. So, we also want:

$$x \ge 2.50$$

Combining these, we find that the solution to the compound inequality is:

$$2.50 \le x \le 4$$

This is an INTERSECTION, or AND problem, because we want both of the requirements to be satisfied.

Solving Intersection Inequality

Let's work through another example

$$-2<x+3<5$$

This inequality doesn't have "OR" written, and has two statements to satisfy. That means we are looking at an "AND" inequality.

Let's solve each part individually

$$\begin{align} x+3&>-2\\ x&>-5\\ \\ x+3&<5\\ x&<2\end{align}$$

We have two separate values that must be satisfied. $x$ has to be less than $2$ and greater than $-2$. Let's look at this on a number line.

Any number in the interval $(-2,2)$ will satisfy both parts of the inequality. We used $<$ instead of $\le$, so $-2$ and $2$ are not included in the interval. Graphically, we can see that the circles are open, so it's not including $2$ and $-2$.

Solving Union Inequality

Some inequalities only require one condition to be fulfilled. Let's look at one.

$$3x+1 \ge 4 \text{ OR } 5x-3 \le -13$$

Just like an intersection, we're going to solve each section one at a time.

$$\begin{align} 3x+1 &\ge 4\\ 3x&\ge 3\\ x&\ge 1\\ \\ 5x-3 \le -13\\ 5x \le -10\\ x \le -2 \end{align}$$

The solution is $x \ge 1 \text{ OR } x \le -2$. Let's look at this on a number line.

Both of the signs in this problem are $\ge$ and $\le$ instead of $>$ and $<$. This means the endpoints are included in the interval. In interval notation, our solution is $(-\infty, -2]\cup[1,\infty)$.