In this lesson we will learn about Compound Inequalities and what it means when we have two or more inequalities with “and/or” statements. We will learn there is a slight difference between the truth of a statement with “and/or” in our spoken language and the truth of a statement with “and/or” in the mathematical language.

Below we will look at the truth of “and/or” statements in our spoken language and mathematical language.

If I said “I am from Oregon and my name is Julian”, for this statement to be true, both parts of the statement would have to be true. If I said “I am from the North Pole and my name is Julian”, even though my name might be Julian, if I am not from the North Pole, then the **whole** statement is false. So, for “and” statements **both** parts of the statement must be true for the **whole** statement to be true.

It follows the __same rules as in our spoken language.__ So, if I said “2 > 1 and 100 > 10” this is a true statement if **both** parts of the statement are true. We know 2 is bigger than 1 and 100 is bigger than 10, so as **both** parts of the statement are true, the **whole** statement is true.

If I said “My name is Julian or I am from Oregon”, and both parts of that statement were true, then in our spoken language we would think that statement is false, as it is like saying “either I am this or I am that”. **“Or” statements are where we have a difference between spoken language and mathematical language**.

“Or” statements in math follow a different rule than in our spoken language. In Mathematical language, an “or” statement is **true as long as one part of the statement is true**. It is **also true if both parts of the statement are true**. If we said “3 > 1 or 99 > 9 ”, it doesn't matter that both parts are true, at least one part of the “or” statement is true, therefore it is mathematically true.

- A compound “and” statement is true if
**all parts of the statement are true**. - A compound “or” statement is true if
**at least one part of the statement is true. It is also true if all parts of the statement are true**.

Note: Throughout this lesson, we will often say “statement” in place of “compound inequality statement” to simplify and abbreviate our working.

Look at the following compound inequality statements and determine if they are mathematically true.

4 < 5 and 7 > -1

For these questions, first, find out what you know about each part of the statement, figure out if each part of the statement is true or false, then look at if it’s an “and” or a “or” statement, then determine if the whole statement is true.

We know that,

4 is less than 5,

7 is greater than -1,

As it is an “and” statement, both parts need to be true for the whole statement to be true,

Both parts of the statement are true, so,

The statement is true.

-6 ≤ -3 or 5 > 104

We know that,

[Don't be thrown off by the ≤ sign, we are still looking at if each part of the statement is true, so we check if -6 is less than or equal to -3]

-6 is less than / equal to -3,

5 **is not** greater than 104,

This is an “or” statement, so **only one part needs to be true**, only one part is true, so,

The statement is true

5 < -2 and 6 ≥ 1

We know that,

5 **is not** less than -2,

6 is greater than / equal to 1,

As this is an “and” statement, **both parts need to be true**. One part is true and one part is false, therefore,

The statement is false.

-2 > -7 or 16 > 12

We know that,

-2 is greater than -7,

16 is greater than 12,

As this is an “or” statement, **only one part needs to be true**, however as we remember for “or” statements, **both parts can be true as well**. We see that both parts are true, therefore,

This statement is true.

7 < -8 or 10 ≥ 11

We know that,

7 is **not less than** 8,

10 is **not less than / equal to** 11,

As this is an “or” statement, **only one part needs to be true**, neither parts of this statement are true, therefore,

This statement is false.

Now, we will move onto checking the truth of inequalities that have algebraic expressions with certain values of variables, let's remind ourselves that of the definition for solutions and solution sets,

The solution / solution set of an Inequality or expression is true if it makes the inequality or expression true. It is not a solution / solution set if it makes the inequality or equation false.

So, we will be checking or solving for variables in inequalities to see if they are true. If they are true, they will be part of the solution set. We will be checking inequalities that involve “and/or” statements, by applying the methods we learned above.

Determine if the given value of x is in the solution set to the compound inequality statements given.

x > -2 and x < 9, with x = 3

Try to not be thrown off by the variable, x, in the statement, think about the methods we have been using and what the question is asking us.

Is x = 3 in the solution set of x > -2 and x < 9?

Remember, if something is **in the solution set** of an inequality (like the one we have above) or an equation, it means **it makes that inequality or equation true**.

So, we want to check if the value x = 3 makes our inequality true,

Does subbing in x = 3 into x > -2 and x < 9 true?

We know how to sub in a value of a variable,

We know how to check if a compound inequality statement is true,

First sub in our value x = 3,

(3) > -2 and (3) < 9,

Now, check if the compound inequality statement is true,

3 is greater than -2,

3 is less than 9,

We have an “and” statement, **so both parts need to be true**, both parts **are** true, so,

The compound inequality statement is true, when x = 3,

As x = 3 makes the statement true, it is therefore **in the solution set**.

x ≤ -9 or x > 9, x = -9

Again, we are checking if x = -9 is in the solution set of our compound inequality statement x ≤ -9 or x > 9,

This means we want to check if subbing x = -9 into x ≤ -9 or x > 9 makes that statement true,

First, we sub in x = -9,

(-9) ≤ -9 or (-9) > 9

Now, we check if that statement, with the x value subbed in, is true,

-9 is less than / equal to -9,

-9 **is not** greater than 9,

This is an “or” statement, so **only one part of the statement needs to be true**, we have one part of the statement that is true, therefore,

The statement is true when x = -9

As the value for the variable x makes the compound inequality statement true, it is therefore in the solution set.

2x + 4 > 7 or 9 - 3x < 4, x = 2

We have slightly longer expressions in our compound inequality statements, but we still follow the same format,

We are checking if x = 2 is in the solution set of 2x + 4 > 7 or 9 - 3x < 4,

So, we are checking that, if we sub x = 2 into 2x + 4 > 7 or 9 - 3x < 4, does it make that compound inequality statement true?

First, we sub in x = 2,

2(2) + 4 > 7 or 9 - 3(2) < 4

Take our time as we calculate the values of each part of the statement,

4 + 4 > 7 or 9 - 6 < 4,

8 > 7 or 3 < 4

Now we have our simplified version of the statement with our value subbed in, check if the statement is true,

8 is greater than 7

3 is less than 4

As this is an “or” statement, we **only need one part to be true for the whole statement to be true**, but **both it is also true if both parts of the statement are true**. We have both parts of the statement as true, therefore,

The statement is true,

So, our statement 2x + 4 > 7 or 9 - 3x < 4 is true when x = 2,

So, when we sub in the value x = 2 into our statement 2x + 4 > 7 or 9 - 3x < 4 it makes the statement true, therefore, x = 2 is part of the solution set.

3x + 2 < 6 and {-3(x - 7)} / 4 = 3, x = 3

As we may notice, instead of two inequalities making up this statement, we have an inequality and an equation, this does not change our approach,

We are checking if x = 3 is part of the solution set of the statement 3x + 2 < 6 and {-3(x - 7)} / 4 = 3,

Which means, we are subbing in x = 3 into 3x + 2 < 6 and {-3(x - 7)} / 4 = 3, and checking if it makes the statement true,

So, with the part with the equation, we are checking if subbing in x = 3 makes that part true, just like we have been with the inequalities.

So, we sub in x = 3 into our statement,

3(3) + 2 < 6 and {-3((3) - 7)} / 4 = 3

Be careful as we multiply out and simplify, there are a few negatives to trip over,

Doing the brackets first,

3(3) + 2 < 6 and {-3(-4)} / 4 = 3

Multiply,

9 + 2 < 6 and 12 / 4 = 3,

11 < 6 and 3 = 3,

Determine if these parts are true,

11 **is not** less than 6,

3 is equal to 3,

This is an and statement, so **we need both parts of the statement to be true**. Only one part is true, therefore,

The statement is false,

As the statement is **false** when **we have subbed in x = 3**, it means that x = 3 is **not in the solution set** of the statement 3x + 2 < 6 and {-3(x - 7)} / 4 = 3

Write in words the solution set to the compound inequalities (this is to give an idea of what the solution set is), then plot the solution set of the compound inequality on a number line.

-1 ≤ x and x < 5

First, we write in words the solution set,

So, we are looking for all the values of x that make the statement true,

We have an “and” statement, so our solution set will be the values that satisfy both parts of the statement,

We need all the values that satisfy -1 ≤ x and x<5,

This means the solution set is all the values that are greater than or equal to -1 **and** less than 5.

This will help us plot the solution set on a number line,

Start by writing two dots above our restricted values, -1 and 5.

As we remember from the previous lesson, when writing solution sets on number lines, if the restricted value is a “greater/less than or equal to” (=< / >=) value, we write a **solid** dot, if it is just a “greater/less than” (< / >) value, we write a hollow dot. So above -1 we write a **solid** dot and above 5 we write a hollow dot.

Now, to finish our number line, we write a line over top of all the values that are in the solution, connecting our two dots

These are all the numbers that are greater than or equal to -1 and less than 5, so we write,

There we have our solution set written on a number line, this gives us a visual of all the values that make our statement -1 ≤ x and x < 5.

Let's repeat this with another example

x ≥ 3 or x < 0

Roughly write what the solution set is in words.

So, we are looking for all the values that are greater than or equal to 3 **or** less than 0.

This means our solution set is all the values that are greater than or equal to 3 **or** less than 0.

For our number line, start by writing a **solid** dot above 3, and a hollow dot above 0.

Now, as we have two parts to this solution set (all the values greater than or equal to 3 or all the values less than 0), we will have two lines,

Also, both parts of this solution set have infinite values,

There is an infinite amount of numbers greater than or equal to 3,

There is an infinite amount of numbers less than 0,

So, both lines drawn on the number line will have arrows at the end to represent that the number line is infinite,

So, let us write our number line over the values that are in our solution set, with an arrow at the end of each line, starting at each of our dots drawn,

Graph the inequality -5/2 < x < 2, try and graph this line without individually drawing the line for x < 2 and x > -5/2

With fractions it can help to write it as a number, so -5/2 becomes -2.5. This can help us to see better where on the number line it will place.

Write our hollow dots above -5/2 and 2,

Now, we see we want the values that are bigger than -5/2 (-2.5) and smaller than 2, so its the values in between those two points, this means our line will connect our two hollow lines we have drawn.