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Sometimes, inequalities can be combined to make a *compound inequality*. This can be done in more than one way, yielding different compound inequalities.

Combining two or more inequalities with the word and

or or

yields what is called a compound inequality.

Compound inequality | Is read as |
---|---|

$x < 5$ or $x > 8$ | $x$ is less than $5$ or greater than $8.$ |

$x > 2$ and $x \leq 4$ | $x$ is greater than $2$ and less than or equal to $4.$ |

Compound inequalities using the word and

are commonly written without the and.

For example,
$x > 2 \ \text{and} \ x \leq 4,$
can be expressed as a compound inequality by rewriting the first inequality as $2 < x$ — the statement $x$ is greater than $2$

is equivalent to $2$ is less than $x.$

The inequality is then
$2 < x \text{ and } x \leq 4,$
which is commonly condensed into

The solution set of a compound inequality consists of the solution sets of the individual inequalities. For compound inequalities with or,

a solution of **either** individual inequality is a solution of the compound inequality. Therefore, the graph of the compound inequality is the union of the graphs of the individual inequalities. These graphs are recognized by the fact that they continue infinitely in either direction.

Combine solution sets

A solution of a compound inequality with and,

however, must be a solution of **both** individual inequalities. Thus, the graph of the compound inequality is the intersection of the graphs of the individual inequalities. These graphs do **not** extend infinitely.

Combine solution sets

Given the following two graphs, find the corresponding compound inequalities.

Show Solution

To begin, notice that the graph does **not** continue infinitely in either direction — it must be the intersection of two inequalities. There is a closed circle at $\text{-} 1,$ and the solution set is to the right of this circle. Thus, $x \geq \text{-} 1$ is one of the two inequalities. Furthermore, the solution set is to the left of the closed circle at $6$ — the second inequality is $x \leq 6.$ Since the graph is the intersection of two inequalities, they are combined with $``$and" to find the compound inequality:
$x \geq \text{-} 1 \quad \text{and} \quad x \leq 6$
or, alternatively, $\text{-} 1 \leq x \leq 6.$

In contrast to the first graph, notice that the second graph continues infinitely in both directions. This means it is the union of two inequalities. Thus, they will be combined using $``$or." There is an open circle at $\text{-} 2,$ from which the solution set extends to the left. Thus, the first inequality is $x < \text{-} 2.$ From the open circle at $5,$ the solution set extends to the right. This represents the second inequality, $x > 5.$ Lastly, combining these two gives us the following compound inequality. $x < 2 \quad \text{or} \quad x > 5$

Solving a compound inequality is done by first separating it into its individual inequalities, which are then solved one at a time as normal. Lastly, their solution sets are combined. As an example, the following inequality can be solved using this method. $\text{-} 3 < 2x - 1 \leq 2$

Separate the compound inequality

Solve the individual inequalities

The individual inequalities can now be solved one at a time. For the example used, $\text{-} 3 < 2x - 1$ will be solved first, using inverse operations.

$\text{-} 3 < 2x - 1$

AddIneq$\text{LHS}+1<\text{RHS}+1$

$\text{-} 3 + 1 < 2x - 1 + 1$

AddTermsAdd terms

$\text{-} 2 < 2x$

DivIneq$\left.\text{LHS}\middle/2\right.<\left.\text{RHS}\middle/2\right.$

$\dfrac{\text{-} 2}{2} < \dfrac{2x}{2}$

SimpQuotSimplify quotient

$\text{-} 1 < x$

The solution set is $\text{-} 1 < x.$ Next is $2x - 1 \leq 2,$ which is solved in a similar manner.

$2x - 1 \leq 2$

AddIneq$\text{LHS}+1\leq\text{RHS}+1$

$2x - 1 + 1 \leq 2 + 1$

AddTermsAdd terms

$2x \leq 3$

DivIneq$\left.\text{LHS}\middle/2\right.\leq\left.\text{RHS}\middle/2\right.$

$\dfrac{2x}{2} \leq \dfrac{3}{2}$

SimpQuotSimplify quotient

$x \leq \dfrac{3}{2}$

The inequality has the solutions $x \leq \frac{3}{2}.$

Combine the solution sets

Solve the following compound inequality and graph the solution set on a number line. $\text{-} 3x + 1 \leq 10 \quad \text{or} \quad \text{-} \dfrac{1}{2}x + 7 \geq 8$

Show Solution

This compound inequality is the combination of two inequalities, using the word $``$or." Thus, we can solve the inequalities individually, and then combine their solution sets using the word $``$or" again. Let's start by solving $\text{-} 3x + 1 \leq 10.$ Remember that when we multiply or divide an inequality by a negative number, the inequality symbol reverses.

$\text{-} 3x + 1 \leq 10$

SubIneq$\text{LHS}-1\leq\text{RHS}-1$

$\text{-} 3x + 1 - 1 \leq 10 - 1$

SubTermsSubtract terms

$\text{-} 3x \leq 9$

DivNegIneqDivide by $\text{-} 3$ and flip inequality sign

$\dfrac{\text{-} 3x}{\text{-} 3} \geq \dfrac{9}{\text{-} 3}$

SimpQuotSimplify quotient

$x \geq \text{-} 3$

The solution set of the first inequality is $x \geq \text{-} 3.$ We'll now solve the second inequality in the same way.

$\text{-} \dfrac{1}{2}x + 7 \geq 8$

Solve for $x$

SubIneq$\text{LHS}-7\geq\text{RHS}-7$

$\text{-} \dfrac{1}{2}x + 7 - 7 \geq 8 - 7$

SubTermsSubtract terms

$\text{-} \dfrac{1}{2}x \geq 1$

MoveRightFacToNumOne$\dfrac{1}{b}\cdot a = \dfrac{a}{b}$

$\text{-} \dfrac{x}{2} \geq 1$

MultIneq$\text{LHS}\cdot 2 \geq \text{RHS} \cdot 2$

$\text{-} \dfrac{x}{2} \cdot 2 \geq 1 \cdot 2$

MultiplyMultiply

$\text{-} x \geq 2$

DivNegIneqDivide by $\text{-}1$ and flip inequality sign

$x \leq \text{-} 2$

Combining this solution set with the one previously found, we get $x \leq \text{-} 2 \quad \text{or} \quad x \geq \text{-} 3.$ This inequality reads $``x$ is less than or equal to $\text{-} 2,$ or greater than or equal to $\text{-} 3."$ Since it's combined with $``$or," its graph is the union of the individual inequalities' graphs. The first, $x \leq \text{-} 2,$ includes $\text{-} 2$ and all numbers smaller than $\text{-} 2.$ Therefore, it is graphed as a closed circle at $\text{-} 2$ with an arrow to the left. Similarly, $x \geq \text{-} 3$ is graphed as a closed circle at $\text{-} 3$ and an arrow to the right. Let's graph these on a number line.

The union of these graphs ends up covering the entire number line. This means that every number is a solution of the compound inequality.

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