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Compound inequalities

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. That means that the final solution will be the union of solutions of separate inequalities. 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 that the final solution will be the intersection of solutions of separate inequalities.

Remember:

or $\Rightarrow$ union

               and $\Rightarrow$ intersection

 

Example with or inequality

$x<5$ 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?

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

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

 

 

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:

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

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$

$ 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 $. Therefore, the solution is the set $ \left<-\infty, – 2\right]$.

Example 2:

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

$57x$ 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 $x > – 5$

$ x < -\frac{1}{3}$ 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$ or $ -x > 7$

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

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:

$\frac{x}{2} > 3$ 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 $\left[- 3, +\infty\right>$.

 

Compound inequalities worksheets

  Integers (946.4 KiB, 694 hits)

  Decimals (1.0 MiB, 417 hits)

  Fractions (1.2 MiB, 417 hits)