PreCalculus I → 2.3 – Sets and Compound Inequalities

2.3 – Sets and Compound Inequalities

Sets, Union and Intersection

A set is a collection of objects called elements. Elements are simply “things” and are not specifically defined. They are known as objects, and are typically mathematical “things” like numbers, but need not be. The months of the year might be elements. The names of people in the class might be the elements of the set, in this case, the class. Simply stated, elements are something that you can put together in a set. Each unique element in the set needs to only be listed once. There should be no repetition on a set.

One way of identifying a set is the Roster Method. This is also called the List method, as the elements are simply listed out. To start, we indicate that we are dealing with a set by using curly brackets { }.

Write the first 4 even numbers as a set
Writing this set in roster method is the easiest. We simply write the numbers inside the curly brackets. $\{2,4,6,8\}$

Write the first 4 even numbers as a set

Writing this set in roster method is the easiest. We simply write the numbers inside the curly brackets.

$\{2,4,6,8\}$

We can identify a set using a capital letter (or lowercase, if desired … or any other symbol really. A capital letter, however, is used most often) Doing so allows us to recall the set without relisting the entire set.

We can also indicate that an element is part of the set by using the symbol “is an element of”: $\in$ . If an element is not part of the set, we use the symbol “not in” $\not\in$ .

A set that has no elements in it is called an empty set, and is denoted with the symbol $\varnothing$

Calling the set $S$ , write the letters of “San Francisco” as a set. Write the letters in alphabetical order in the set, and then indicate that the letter f is an element of the set and that z is not. Ignore capitalization of elements
$S=\{\text{a, c, f, i, n, o, r, s}\}$ Note that we only included each letter once, regardless of the number of times it appeared in “San Francisco”. To indicate that f is an element of the set $S$ , we write: $\text{f}\in\ S$ To indicate that z is not an element of the set $S$ , we write $\text{z}\not\in S$

Calling the set $S$ , write the letters of “San Francisco” as a set. Write the letters in alphabetical order in the set, and then indicate that the letter f is an element of the set and that z is not. Ignore capitalization of elements

$S=\{\text{a, c, f, i, n, o, r, s}\}$ Note that we only included each letter once, regardless of the number of times it appeared in “San Francisco”. To indicate that f is an element of the set $S$ , we write:

$\text{f}\in\ S$ To indicate that z is not an element of the set $S$ , we write

$\text{z}\not\in S$

Write the set of numbers that are both positive and negative.
Because no number is both positive and negative, we have no elements in the set. Thus, we have the empty set. $\varnothing$

We will discuss two ways that sets can be joined together. The first is called a union. A Union of two sets is a new set that is composed of elements of either one or the other, or both sets. If we wish to find the union of set $A$ and set $B$ , we notate this as $A\cup B$

Find the union of the two sets $A$ and $B$ given below:
$A=\{1,3,6,7,9\}$ $B=\{2,3,4,5,9\}$ The union of these two sets is the new set whose elements are all of those from set $A$ combined with all of the elements from $B$ . any duplicates are only written once. For this union, we have: $A\cup B=\{1,2,3,4,5,6,7,9\}$

Given the following sets, find the union $F\cup G$
$F=\{a, f, t, y, z\}$ $G=\{a, d, f, z\}$ For this union, we have: $F\cup G=\{a, d, f, t, y, z\}$

The second method of combining sets is in an intersection. An Intersection of two sets is a new set composed of all elements that are in both sets. We notate the intersection of sets $A$ and $B$ as: $A\cap B$

For the given sets, find $A\cap B$
$A=\{1, 2, 4, 7, 8, 15\}, \ \ \ \ \ B=\{3, 5, 7, 15\}$ The intersection $A\cap B$ is only the elements that are in both sets $A$ and $B$ . Thus, we have: $A\cap B = \{7, 15\}$

For the given sets, find $A\cap B$

$A=\{1, 2, 4, 7, 8, 15\}, \ \ \ \ \ B=\{3, 5, 7, 15\}$ The intersection $A\cap B$ is only the elements that are in both sets $A$ and $B$ . Thus, we have:

$A\cap B = \{7, 15\}$

Find the intersection of the given sets
$N=\{1,2,3,a,b,c\}$ $M=\{1,3,6,b,h,i\}$ The intersection is $N\cap M=\{1,3,b\}$

Find both the union and the intersection of the sets
$A=\{2, 3, 4, 5\}$ $B=\{1, 3, 5, 7, 10\}$ For each, we have: $\begin{align} A\cup B&=\{1, 2, 3, 4, 5, 7, 10\}\\ A\cap B&=\{3, 5\} \end{align}$

Find both the union and the intersection of the sets

$A=\{2, 3, 4, 5\}$
$B=\{1, 3, 5, 7, 10\}$ For each, we have:

$\begin{align*} A\cup B&=\{1, 2, 3, 4, 5, 7, 10\}\\ A\cap B&=\{3, 5\} \end{align*}$

Inequalities share a property with sets in that inequalities can be combined together in the equivalence of unions and intersections to form compound inequalities. When working with inequalities, we have an “OR” combination and an “AND” combination. These are called Compound Inequalities.

An OR compound inequality will result in a true statement from either one inequality, the other, or both. This type of inequality is called a disjunction, and the solution of the compound inequality is the union of the solution sets of the individual inequalities.

To easily graph the union of the sets, it is easy to graph the individual solutions of each inequality, and then combine them into a single solution.

Find the solutions to the compound inequality. Graph the solution and write the solution in interval notation.
$x>5 \ \ \ \text{or} \ \ \ x<-3$ We can graph each individual solution, The graph of $x>5$ on top and the graph of $x<-3$ directly underneath: And when we combine the solutions that are on either one or the other solution set above, we have: The interval notation for this solution is a union of the two solutions. We write this as $(-\infty,-3)\cup(5,\infty)$

Find the solutions to the compound inequality. Graph the solution and write the solution in interval notation.

$x>5 \ \ \ \text{or} \ \ \ x<-3$ We can graph each individual solution, The graph of $x>5$ on top and the graph of $x<-3$ directly underneath:

And when we combine the solutions that are on either one or the other solution set above, we have:

The interval notation for this solution is a union of the two solutions. We write this as

$(-\infty,-3)\cup(5,\infty)$

Find the solution of the compound inequality. Graph the solution and write the solution in interval notation.
$3x-4\ge5 \ \ \ \text{or} \ \ \ 2x+5 <3$ For this compound inequality, we must solve each inequality separately: $3x-4\ge 5 \hspace{0.2cm} or \hspace{0.2cm} 2x+5<3$ $3x\ge9 \hspace{0.2cm} or \hspace{0.2cm} 2x<-2$ $x\ge3 \hspace{0.2cm} or \hspace{0.2cm} x<-1$ Graphing the solution is again easy if we graph the solutions to each inequality individually: And putting the two together, we have: And to write the solution in interval notation: $(-\infty,-1)\cup [3,\infty)$

Find the solution of the compound inequality. Graph the solution and write the solution in interval notation.

$3x-4\ge5 \ \ \ \text{or} \ \ \ 2x+5 <3$

For this compound inequality, we must solve each inequality separately:

$3x-4\ge 5 \hspace{0.2cm} or \hspace{0.2cm} 2x+5<3$

$3x\ge9 \hspace{0.2cm} or \hspace{0.2cm} 2x<-2$

$x\ge3 \hspace{0.2cm} or \hspace{0.2cm} x<-1$ Graphing the solution is again easy if we graph the solutions to each inequality individually:

And putting the two together, we have:

And to write the solution in interval notation:

$(-\infty,-1)\cup [3,\infty)$

An AND compound inequality will result in a true statement for values that are a solution to both equalities. The solution to these inequalities is the intersection of the two individual solutions. To graph them, we will graph the individual solutions again, and then express the intersection on its own graph.

Find the solutions to the compound inequality. Graph the solution and write the solution in interval notation
$x>-2 \ \ \ \text{and} \ \ \ x<4$ We first graph each inequality separately: And the solution to the compound inequality is only the values that are a solution to both, so we have our final solution: And to write the intersection in interval notation, we need only express one interval, so we have: $(-2,4)$

Find the solutions to the compound inequality. Graph the solution and write the solution in interval notation

$x>-2 \ \ \ \text{and} \ \ \ x<4$

We first graph each inequality separately:

And the solution to the compound inequality is only the values that are a solution to both, so we have our final solution:

And to write the intersection in interval notation, we need only express one interval, so we have:

$(-2,4)$

Find the solutions to the compound inequality. Graph, and express the solution in interval notation.
$3x+7\ge16 \ \ \ \text{and} \ \ \ 2x-7\le3$ We will need to solve each inequality: $3x+7\ge16 \hspace{0.2cm} \text{and}\hspace{0.2cm} 2x-7\le3$ $3x\ge9 \hspace{0.2cm} \text{and}\hspace{0.2cm} 2x\le10$ $x\ge3 \hspace{0.2cm} \text{and}\hspace{0.2cm} x\le5$ and graph each inequality. And the intersection of the above graphs is shown here: Then the interval notation of the solution is $[3,5]$

Find the solutions to the compound inequality. Graph, and express the solution in interval notation.

$3x+7\ge16 \ \ \ \text{and} \ \ \ 2x-7\le3$

We will need to solve each inequality:

$3x+7\ge16 \hspace{0.2cm} \text{and}\hspace{0.2cm} 2x-7\le3$

$3x\ge9 \hspace{0.2cm} \text{and}\hspace{0.2cm} 2x\le10$

$x\ge3 \hspace{0.2cm} \text{and}\hspace{0.2cm} x\le5$ and graph each inequality.

And the intersection of the above graphs is shown here:

Then the interval notation of the solution is $[3,5]$

Find the solutions to the compound inequality. Graph, and express the solution in interval notation.
$2(x+2)+5>3 \ \ \ \text{and} \ \ \ -3(x+1)\ge-9$ Solve each inequality: $2(x+2)+5>3 \ \ \ \text{and} \ \ \ -3(x+1)\ge-9$ $2x+4+5>3 \ \ \ \text{and} \ \ \ -3x-3\ge -9$ $2x+9>3 \ \ \ \text{and} \ \ \ -3x-3\ge -9$ $2x>-6 \ \ \ \text{and} \ \ \ -3x \ge -6$ $x>-3 \ \ \ \text{and} \ \ \ x{\color{red}\le}2$ Graphing each inequality, we have: The intersection is the values that are solutions to BOTH inequalities. Note that the value $-3$ is NOT a solution to the first inequality, and thus it is not in the intersection. Thus, it is also an open circle in the intersection: The interval notation solution is $(3,2]$

Find the solutions to the compound inequality. Graph, and express the solution in interval notation.

$2(x+2)+5>3 \ \ \ \text{and} \ \ \ -3(x+1)\ge-9$

Solve each inequality:

$2(x+2)+5>3 \ \ \ \text{and} \ \ \ -3(x+1)\ge-9$

$2x+4+5>3 \ \ \ \text{and} \ \ \ -3x-3\ge -9$

$2x+9>3 \ \ \ \text{and} \ \ \ -3x-3\ge -9$

$2x>-6 \ \ \ \text{and} \ \ \ -3x \ge -6$

$x>-3 \ \ \ \text{and} \ \ \ x{\color{red}\le}2$ Graphing each inequality, we have:

The intersection is the values that are solutions to BOTH inequalities. Note that the value $-3$ is NOT a solution to the first inequality, and thus it is not in the intersection. Thus, it is also an open circle in the intersection:

The interval notation solution is $(3,2]$

Tripartite

A compound inequality consisting of a single line and two inequality symbols is called a tripartite. An example of a tripartite inequality is

$-4<x\le10$
which means the compound inequality:

$-4<x$ and $x\le10$ We can work with a tripartite inequality similarly to working with an single inequality. In order to isolate the variable, we will add, subtract, multiply and divide, but we must ensure that whatever is done to one part of the tripartite is done to all three parts!

Express the solution set of the compound inequality $2<x\le5$
We see that numbers between two and 5, including 5 will solve the inequality, so our solution set in interval notation is $(2,5]$ . We can also write this in set builder notation. This would be: $\{x\|2<x\le5\}$ . We might also graph this:

Express the solution set of the compound inequality $2<x\le5$

We see that numbers between two and 5, including 5 will solve the inequality, so our solution set in interval notation is $(2,5]$ . We can also write this in set builder notation. This would be: $\{x|2<x\le5\}$ . We might also graph this:

Express the solution set to the inequality $-8< 4x < 16$ in interval notation
We must first solve this inequality for $x$ . We can use the multiplication principle, as long as we multiply or divide all sides of the inequality by the same number: $\frac{-8}{\color{red}4}<\frac{4x}{\color{red}4}<\frac{16}{\color{red}4}$ $-2<x<4$ So our solution set is: $(-2,4)$

Express the solution set to the inequality $9> -2x+5 > -13$ in interval notation.
Again, we must solve the compound inequality for $x$ . $9{\color{red}-5}>-2x+5{\color{red}-5}>-13{\color{red}-5}$ $4>-2x>-18$ $\frac{4}{\color{red}-2}>\frac{-2x}{\color{red}-2}>\frac{-18}{\color{red}-2}$ Here, we can not forget to flip both inequalities because we are dividing by a negative: $-2<x<9$ So then our solution set is (-2,9).

PreCalculus I

2.3 – Sets and Compound Inequalities

Sets, Union and Intersection

Tripartite

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