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What are compound inequalities and difference from simple inequalities

compound inequalities

Compound inequalities are inequalities bounded by more than one condition. What does it mean? Simple inequalities have one unknown, certain value and an inequality sign between them. In compound inequalities there are two inequality signs. This may seem complicated, but there is no reason to worry. We can simply break down one inequality into two simple ones which know to solve. But, we need to be careful…

In this lesson, there are basically only two different type of problems: one in which the conditions are bounded with an operation ‘AND’ and second in which the conditions are bounded with an ‘OR’.

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. When you will be working with inequalities you will always get a set of numbers as a solution.

They will usually come in the following form: condition1 <(≤) x <(≤) contition2, which can be w broken down into two simple inequalities: condition1 <(≤) x AND x <(≤) contition2.

Example 1.

5 < x < 9
This is one type of inequality. If we want to solve it, first we have to break it down into two separate inequalities. For the first inequality you’ll take whole left side, first inequality sign and go all the way to the next inequality sign. For the second inequality you’ll take the part from the first inequality sign to the end.

Which means that now you have: 5 < x and x < 9. Solve them separately, write down their solutions and draw them on the number line.

1. 5 < x => x > 5 => x e <5, +\infty>

2. x < 9 => x e <-∞, 9>

x is bigger than 5 and less than 9

For every compound inequality in this form the solution will be the common solution of the inequalities you got by dividing it. Always draw your solutions on the number line; that is the best way for you to see the solution. The solution will be the part of a number line where the lines of solutions overlap.

In this example our solution will be <5, 9>. Also don’t forget the difference between different brackets and if you are considering the endpoints as parts of the solution or not.

solution last example

Now take a look at a bit more complicated example 2:

2 + 2x \le x < 5 + x
This is solvable in two ways.

First is the same way you did the last example. Divide into two inequalities and solve separately, the solution will again be the intersection between the sets of solutions of those two inequalities.

2 + 2x < x => 2 < -x => x < - 2 -> x e <-\infty, - 2]

x < 5 + x => 0 < 5
In second inequality we ‘lost’ our unknown. We got 0 < 5 which is a statement that is always true. This means that no matter which number you put instead of the unknown in expression x < 5 + x it will be true, and that the solution is the whole set of real numbers. In case you got something that is not true, for example 5 < 0, that inequality will have no solutions.

If we draw those solutions on a number line we get:

negative infinity to minus 2

So our solution is the set <-\infty, - 2].

Now, if you are already comfortable with inequalities you can do the following and solve it a little bit faster. If you have a inequality like this, and want to do some operations on it, you have to do it for every part, for example if you’re going to add something on the left, you have to add it in the middle and on the right. If you noticed at first that you have an always true statement you can subtract from every part -2x and then you’ll have

2 \le - x < 5 - x

- x < 5 - x is again always true statement, so we only consider 2 \le −x.

You can also use this procedure to simplify, if you can, your inequalities. For example if you have something like example 3.

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

If you subtract from every part 57x you get 1 + x < - 2x < 10.

And then you simply continue with the procedure you learned before.

1 + x < - 2x and - 2x < 10 (here we came across a minus, if you are multiplying or dividing with minus the inequality sign must change)

3x < - 1 and x > - 5

x < -\frac{1}{3} and x > - 5

solution is between -5 and 1/3

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

In case your lines don’t overlap your inequality will have no solutions.

For example if you get x > 5 and x > - 7.

Also, you can get different forms of inequalities.

Example 4.

5 > x > 2

You solve this in the same way. 5 > x and x > 2x e <5, 2>.

Example 5.

- 6 < x > 2

x > - 6 and x > 2 => x e <2, +\infty>

There are also compound inequalities whose conditions are bounded with an ‘or’. The only difference between solving them and solving those with an ‘and’ is that in your final solution you will look for a union of separate inequalities.

Example 6.

2x > 4 or -x > 7

2x > 4 -> x > 2

- x > 7 => x < - 7
solution compound inequalities

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

Example 7.

\frac{x}{2} > 3 or x \ge  - 3

x > 6 or x \ge  - 3

solution in example

In cases like this, you don’t have to write to separate sets because one contains another. The solution will be the greater one in this case the solution is [- 3, +\infty>.


Compound inequalities worksheets

  Integers (946.4 KiB, 497 hits)

  Decimals (1.0 MiB, 298 hits)

  Fractions (1.2 MiB, 301 hits)