A **compound** **inequality**, sometimes referred to as **combined inequality**, is an inequality that combines two or more simple inequalities joined together with ** or **or

**.**

*and*To be a solution of an ** or **inequality, a value has to make only one part of the inequality true. That means that the final solution will be the

**union**of solutions of separate inequalities. To be a solution of an

**inequality, it must make both parts true. Inequalities whose conditions are bounded with**

*and**and*are not independent of each other. That means that the final solution will be the

**intersection**of solutions of separate inequalities.

**Remember:**

**or $\Rightarrow$ union**

** and $\Rightarrow$ intersection**

**Example with or inequality**

$x<5 \Rightarrow x \in \left<-\infty, 5\right>$

$x>9 \Rightarrow x \in \left<9, \infty\right>$

The answer are all numbers less than 5 and all numbers greater than 9, so the solution is a union of two intervals $\left<-\infty, 5\right> \cup \left<9, \infty\right>$.

**Example with and inequality**

$x>4$ and $x<7$

This statement is equivalent to $4<x<7$.

$x>4 \Rightarrow x \in \left<4, \infty\right>$

$x<7 \Rightarrow x \in \left<-\infty, 7\right>$

The solution is an intersection of two intervals $\left<-\infty, 7\right> \cap \left<4, \infty\right>$, which is the interval $\left<4, 7\right>$.

**Example 1:**

**I.**

The problem is divided into two inequalities which are then solved **separately**. The solution is the **intersection** of the individual solutions.

If the case was different and the statement wasn’t true, for example $5 < 0$, then the inequality wouldn’t have solutions. For example, the inequality $x<x-1$ has no solutions.

The solution of the inequality from the Example 1 is the set $ \left<-\infty, – 2\right]$.

**II.**

When working with equations, one can add, subtract, multiply and divide the expression, but what is changed on one side must be also changed the same way on the other. It is similar when working with inequalities. If you add, subtract, multiply or divide, you must do it for every part of the inequality.

In our example we can subtract $2x$.

$ 2 + 2x \leqslant x < 5 + x$

$ – x < 5 – x$ is always a true statement, so the only part which restricts the set of solutions is $ 2 \leqslant – x $. Therefore, the solution is the set $ \left<-\infty, – 2\right]$.

**Example 2:**

$57x$ is subtracted from every part of the inequality:

$ 1 + x < – 2x < 10$

$ 1 + x < – 2x$ and $ – 2x < 10$

When the expression $ – 2x < 10$ is divided by $-2$, the sign of inequality changes.

$ 3x < – 1$ and $x > – 5$

The final solution is the set $ \left<- 5, -\frac{1}{3}\right>$.

In case when the intersection is empty, there are no solutions.

For example, if $ x > 5$ and $ x < – 7$, then the intersection of the sets $\left<5, \infty\right>$ and $\left<-\infty, -7\right>$ is empty.Therefore, inequality has no solutions.

**Example 3:**

$ 5 > x > 2$

$ 5 > x$ and $ x > 2 \Rightarrow x \in \left<2, 5\right>$

**Example 4:**

$ – 6 < x > 2$

$ x > – 6$ and $ x > 2 \Rightarrow x \in \left<2, +\infty\right>$

**Example 5:**

$ 2x > 4 \Rightarrow x > 2$

The solution of the first inequality is set $\left<2, \infty\right>$ and the solution of the second inequality is set $\left<-\infty, – 7\right>$.

The final solution is the union of these two sets: $ \left<-\infty, – 7\right> \cup \left<2, \infty\right>$.

**Example 6:**

$ x > 6$ or $ x \geqslant- 3$

## Compound inequalities worksheets

**Integers** (946.4 KiB, 838 hits)

**Decimals** (1.0 MiB, 512 hits)

**Fractions** (1.2 MiB, 511 hits)