Math Core

A thermostat keeps a room between $68$ and $72$ degrees. A carnival ride turns away anyone shorter than $42$ inches or taller than $76$ inches. Each of these rules needs two inequalities working together. Joining two inequalities with "and" or "or" makes a compound inequality.

"And" versus "or"

Definition

Compound inequality

A compound inequality is two inequalities joined by the word and or the word or.

An and inequality is true only when both parts are true. Its solutions are the numbers the two parts have in common (their intersection).
An or inequality is true when at least one part is true. Its solutions are all the numbers that satisfy either part (their union).

Think of the thermostat. The temperature $t$ must satisfy $t \ge 68$ and $t \le 72$ . A temperature of $70$ passes both tests. A temperature of $75$ passes the first but fails the second, so it doesn't work.

An "and" inequality whose solutions lie between two numbers is usually written as one chain:

68 \le t \le 72.

Read it from the middle out: " $t$ is at least $68$ and at most $72$ ." The variable sits in the middle, and both symbols point the same way.

Now think of the ride. A height $h$ is turned away if $h < 42$ or $h > 76$ . A height of $30$ makes the first part true ( $30 < 42$ ) even though the second part is false, and that's enough; one true part makes the whole statement true. An "or" inequality can't be written as a chain, because no number is both less than $42$ and greater than $76$ .

Graphing compound inequalities

An "and" inequality between two numbers graphs as a segment. Here is $-1 \le x < 3$ : a closed circle at $-1$ (included), an open circle at $3$ (not included), and shading between them.

-1 ≤ x < 3

An "or" inequality like $x < -2$ or $x \ge 1$ usually graphs as two rays pointing away from each other, with a gap in the middle. On a single number line you shade both rays. Here are its two pieces, one at a time:

first piece: x < -2

second piece: x ≥ 1

The full solution set is everything shaded in either picture: all numbers left of $-2$ together with $1$ and everything to its right. Numbers in the gap, such as $0$ , are not solutions: $0 < -2$ is false and $0 \ge 1$ is false.

Reading compound inequalities

And (between): $a < x < b$ graphs as a segment from $a$ to $b$ . A solution must pass both tests.
Or (outside): $x < a$ or $x > b$ graphs as two rays going outward. A solution must pass at least one test.

Use closed circles for $\le$ and $\ge$ , and open circles for $<$ and $>$ , exactly as before.

Solving "and" inequalities

When an "and" inequality is written as a chain, it has three parts. Whatever you do to one part, do to all three. The goal is to get the variable alone in the middle.

Worked example: Solving a chain

Solve $-3 \le 2x + 1 < 9$ and graph the solutions.

\begin{aligned} -3 &\le 2x + 1 < 9 \\ -4 &\le 2x < 8 && \text{subtract } 1 \text{ from all three parts} \\ -2 &\le x < 4 && \text{divide all three parts by } 2 \end{aligned}

-2 ≤ x < 4

Check. At $x = -2$ : $2(-2) + 1 = -3$ , and $-3 \le -3 < 9$ is true, so $-2$ is included. At $x = 4$ : $2(4) + 1 = 9$ , and $9 < 9$ is false, so $4$ is excluded. At $x = 0$ : $-3 \le 1 < 9$ , true.

The same rule about negatives applies: if you divide by a negative number, reverse both symbols.

Worked example: Dividing a chain by a negative

Solve $-5 < 1 - 3x \le 10$ .

\begin{aligned} -5 &< 1 - 3x \le 10 \\ -6 &< -3x \le 9 && \text{subtract } 1 \\ 2 &> x \ge -3 && \text{divide by } -3 \text{ and reverse both symbols} \end{aligned}

The chain $2 > x \ge -3$ is correct but reads backward. Rewrite it with the smaller number on the left: $-3 \le x < 2$ .

Check. $x = 0$ gives $1 - 0 = 1$ , and $-5 < 1 \le 10$ is true. $x = 2$ gives $1 - 6 = -5$ , and $-5 < -5$ is false, so $2$ is excluded, matching the open circle.

Common mistake

After dividing a chain by a negative, write your final answer with the numbers in increasing order, left to right, just as they appear on the number line. A chain like $2 > x \ge -3$ is correct, but it is easy to misread and to graph backward. Also, never write a chain like $5 < x < 1$ : no number is both greater than $5$ and less than $1$ .

Solving "or" inequalities

Solve each part on its own, then join the answers with "or."

Worked example: Solving an or inequality

Solve $2x - 1 < -5$ or $3x + 2 \ge 11$ .

First part: $2x < -4$ , so $x < -2$ .
Second part: $3x \ge 9$ , so $x \ge 3$ .

The solution is $x < -2$ or $x \ge 3$ : two rays, an open circle at $-2$ shaded left and a closed circle at $3$ shaded right.

Check. $x = -4$ : $2(-4) - 1 = -9 < -5$ , so the first part is true and $-4$ is a solution. $x = 0$ : $-1 < -5$ is false and $2 \ge 11$ is false, so $0$ is not a solution.

Special cases

Compound inequalities can have every number or no number as a solution.

Worked example: All or nothing

$x < 5$ or $x > 1$
$x \le 1$ and $x > 4$

Solutions.

Every real number is less than $5$ or greater than $1$ (many are both). Try $100$ : it's greater than $1$ . Try $-100$ : it's less than $5$ . The solution is all real numbers.
A number can't be at most $1$ and also greater than $4$ . The two rays don't overlap, so there is no solution.

Tip

To check any compound answer, test one number in each region the boundaries create. For $x < -2$ or $x \ge 3$ , test something below $-2$ , something between $-2$ and $3$ , and something above $3$ . Only the regions that make the original compound inequality true should be shaded.

Practice

Practice 1

Which inequality says " $x$ is greater than $-4$ and at most $2$ "?

Practice 2

Which inequality is shown by the graph?

Practice 3

The solution of $3 < x + 5 \le 11$ has the form $a < x \le b$ . Enter $a$ and $b$ , smaller first.

Separate answers with commas, e.g. 2, -5

Practice 4

Solve $-7 \le 3x - 1 < 8$ .

Practice 5

Solve $-4 < 2 - 2x < 6$ . The answer can be written as $a < x < b$ . Enter $a$ and $b$ , smaller first.

Separate answers with commas, e.g. 2, -5

Practice 6

Solve $x + 4 < 1$ or $2x - 3 > 7$ .

Practice 7

Which compound inequality has no solution?

Practice 8

Jordan's first three test scores are $78$ , $85$ and $91$ . To earn a B, the average of all four tests must be at least $80$ but less than $90$ . What is the lowest score Jordan can get on the fourth test and still earn a B?

Enter a number. Fractions like 3/4 and sqrt(2) are OK.