# Compound inequalities

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.  To be a solution of an and inequality, it must make both parts true. Inequalities whose conditions are bounded with and are not independent of each other. That means the final solution will be the intersection of solutions of separate inequalities.

Example with or inequality

$x<5$ or $x>9$
The question asked in the example is: which numbers can be substitued in the place of $x$ so that one of the inequalities is true?

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

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

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

Example with and inequality

$x>4$ and $x<7$

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

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

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

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

Example 1.

$2 + 2x \leqslant x < 5 + x$
This problem can be solved in two ways.

I.

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

$2 + 2x \leqslant x$ and $x < 5 + x$
$2 + 2x \leqslant x \Rightarrow 2 < -x \Rightarrow x < – 2\Rightarrow x \in <-\infty, – 2]$
$x < 5 + x \Rightarrow 0 < 5$
From the second inequaltiy follows the statement $0 < 5$ which is always true, so the solution the whole set of real numbers.

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

The solution in the example 1. is the set $<-\infty, – 2]$.

II.

When working with equalities 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 inequalitiy.

In our example we can subtract $2x$.

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

$2 \leqslant – x < 5 – x$

$– x < 5 – x$ is always a true statement, so the only part which restricts the set of solutions is $2 \leqslant – x$.

Example 2.

$1 + 58x < 55x < 57x + 10$

From every part of the inequality $57x$ is subtracted.

$1 + x < – 2x < 10$

$1 + x < – 2x$ and $– 2x < 10$

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

$3x < – 1$ and $x > – 5$

$x < -\frac{1}{3}$ and $x > – 5$

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

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 $<5, \infty>$ and $<-\infty, -7>$ is empty.

Example 3.

$5 > x > 2$

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

Example 4.

$– 6 < x > 2$

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

Example 5.

$2x > 4$ or $-x > 7$

$2x > 4 \Rightarrow x > 2$
$– x > 7 \Rightarrow x < – 7$

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

Example 6.

$\frac{x}{2} > 3$ or $x \geqslant – 3$

$x > 6$ or $x \geqslant- 3$
In a case like this, there is no need to write a union of two sets since one is the subset of the other. The solution is simply the bigger set, in this case $[- 3, +\infty>$.

## Compound inequalities worksheets

Integers (946.4 KiB, 645 hits)

Decimals (1.0 MiB, 387 hits)

Fractions (1.2 MiB, 379 hits)