Given A={2,4,6,8}A=\left\lbrace2,4,6,8\right\rbrace, B={4,8,12,16}B=\left\lbrace4,8,12,16\right\rbrace, C={8}C=\left\lbrace8\right\rbrace, find the following:(B) A∪BA\cup B

{ 2 , 4 , 6 , 8 , 12 , 16 } \left\lbrace2,4,6,8,12,16\right\rbrace { 2 , 4 , 6 , 8 , 12 , 16 }

Given A = { 2 , 4 , 6 , 8 } A=\left\lbrace2,4,6,8\right\rbrace , B = { 4 , 8 , 12 , 16 } B=\left\lbrace4,8,12,16\right\rbrace , C = { 8 } C=\left\lbrace8\right\rbrace , find the following: (B) A ∪ B A\cup B

In this problem, we are given that set A is equal to two, four, six, eight, set B is equal to four, eight, twelve, and sixteen, and set C is equal to just eight. Now we're asked to find the following, and for part b of this example or this problem, we have A union b. So how exactly do I find this? Well, there are two sets that I'm interested in looking at. I'm interested in looking at A and I'm interested in looking at b. Now looking at the two sets which I am given above, I can list out the values that we have in each one. Now I'm gonna start with what we have in a. Notice we have two, and then I also see that we have four, six, and eight. But what I'm gonna do is just write that six. Now there's a reason I'm only writing that six, because notice that the four and the eight also show up in set b, so I'm gonna write the four and the eight in this overlap. Now next I will move to b. For b, well, we already have this four and this eight in common in both of these, and then we uniquely have twelve and sixteen in set b. So we have set b, we have set a, and then four and eight would be the overlap between these sets. Now we're looking for a union here, and the union means we're interested in including all the values that we have between a and b. So because I'm trying to cover all of these values here, well what that means is that my union is just going to be a collection of all these numbers, so we're going to have the set two and six, and actually I'm going to put these in ascending order so we're going to go two and then we have four and then we have six then we have eight, and then ascending order would be then 12, and then 16. Notice you don't want to write the repeating numbers more than once, so we have all of our values here, and that would be the solution and the union of a and b. So hope you found this video helpful, and let's move on and get some more practice with this concept.

Given A = { 2 , 4 , 6 , 8 } A=\left\lbrace2,4,6,8\right\rbrace , B = { 4 , 8 , 12 , 16 } B=\left\lbrace4,8,12,16\right\rbrace , C = { 8 } C=\left\lbrace8\right\rbrace , find the following: (B) A ∪ B A\cup B

{ 2 , 4 , 6 , 8 , 12 , 16 } \left\lbrace2,4,6,8,12,16\right\rbrace { 2 , 4 , 6 , 8 , 12 , 16 }

{ 4 , 8 } \left\lbrace4,8\right\rbrace { 4 , 8 }

{ 2 , 6 , 12 , 16 } \left\lbrace2,6,12,16\right\rbrace { 2 , 6 , 12 , 16 }

2. Linear Equations and Inequalities

Set Operations and Compound Inequalities

Intermediate Algebra

2. Linear Equations and Inequalities

Set Operations and Compound Inequalities

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Multiple Choice

Given $A = {2, 4, 6, 8}$ , $B = {4, 8, 12, 16}$ , $C = {8}$ , find the following:
(B) $A \cup B$

$\left\lbrace2,4,6,8,12,16\right\rbrace$

$\left\lbrace4,8\right\rbrace$

$\left\lbrace2,6,12,16\right\rbrace$

$\phi$

Verified step by step guidance

Identify the two sets given: $A = \left\lbrace 2,4,6,8 \right\rbrace$ and $B = \left\lbrace 4,8,12,16 \right\rbrace$.

Recall that the union of two sets, denoted $A \cup B$, includes all elements that are in $A$, or in $B$, or in both.

List all elements from set $A$: \$2, 4, 6, 8$.

Add all elements from set $B$ that are not already in $A$: \$12, 16$.

Combine these elements into one set without duplicates: $\left\lbrace 2,4,6,8,12,16 \right\rbrace$.

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