Truth Values of and/or
Solution sets of equations and inequalities
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.
An “and” statement in our spoken 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.
An “and” statement in Mathematical Language
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.
An “or” statement in our spoken language:
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.
An “or” statement in 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.
Question 1
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.
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
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.
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.
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.
Question 2
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.
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