Students use the numbers in this three-set Venn diagram to solve complex set theory equations listed below.

Let’s solve each problem step-by-step using the given Venn diagram.

First, let’s understand the diagram:

We have three sets: A, B, and C.

The numbers in the regions represent elements (or counts) in those specific regions.

Let’s label the regions for clarity:

- Only A: 14, 8
- Only B: 5
- Only C: 10, 7, 18
- A ∩ B only (not C): 3, 16
- B ∩ C only (not A): 12, 13
- A ∩ C only (not B): 17, 15
- A ∩ B ∩ C: 9
- Outside all: none shown (but we’ll assume universe is just these numbers unless stated)

So, total elements in each set:

- A = {14, 8, 3, 16, 17, 15, 9} → 7 elements
- B = {3, 16, 9, 12, 13, 5} → 6 elements
- C = {17, 15, 9, 12, 13, 10, 7, 18} → 8 elements

But actually, since we’re dealing with sets of numbers, we treat each number as a distinct element.

So the universal set U = {3, 5, 7, 8, 9, 10, 12, 13, 14, 15, 16, 17, 18}

Now, let’s solve each problem.

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1) A' ∪ (B ∩ C)'

Recall: A' = everything not in A
(B ∩ C)' = everything not in (B ∩ C)

B ∩ C = elements in both B and C = {9, 12, 13} (since 9 is in all three, 12 and 13 are in B and C but not A)

So (B ∩ C)' = U - {9,12,13} = {3,5,7,8,10,14,15,16,17,18}

A' = U - A = U - {3,8,9,14,15,16,17} = {5,7,10,12,13,18}

Now, A' ∪ (B ∩ C)' = union of those two sets.

{5,7,10,12,13,18} ∪ {3,5,7,8,10,14,15,16,17,18} = {3,5,7,8,10,12,13,14,15,16,17,18}

That’s everything except 9? Wait, 9 is missing?

Wait — check: 9 is in A and in B∩C, so it’s not in A' and not in (B∩C)', so it should be excluded.

But let’s verify:

Is there any element missing from the union?

U = {3,5,7,8,9,10,12,13,14,15,16,17,18}

Our union has all except 9.

So answer = U - {9} = {3,5,7,8,10,12,13,14,15,16,17,18}

But perhaps we can write it as “all elements except 9”.

Alternatively, note that by De Morgan’s law:

A' ∪ (B ∩ C)' = (A ∩ B ∩ C)' — wait, no.

Actually: (B ∩ C)' = B' ∪ C'

So A' ∪ (B' ∪ C') = A' ∪ B' ∪ C' = (A ∩ B ∩ C)' — yes!

Because (A ∩ B ∩ C)' = A' ∪ B' ∪ C'

So A' ∪ (B ∩ C)' = A' ∪ B' ∪ C' = (A ∩ B ∩ C)'

A ∩ B ∩ C = {9}, so (A ∩ B ∩ C)' = U - {9} = {3,5,7,8,10,12,13,14,15,16,17,18}

✔ Answer 1: {3,5,7,8,10,12,13,14,15,16,17,18}

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2) (B ∪ C)' ∪ A

First, B ∪ C = elements in B or C = {3,5,9,12,13,15,16,17,10,7,18}

So (B ∪ C)' = U - (B ∪ C) = elements not in B or C = {8,14} (only in A and not in B or C)

Then (B ∪ C)' ∪ A = {8,14} ∪ A = {8,14} ∪ {3,8,9,14,15,16,17} = {3,8,9,14,15,16,17}

Which is exactly set A.

Because (B ∪ C)' is disjoint from B and C, and A may overlap, but since (B ∪ C)' ⊆ A^c, and we’re unioning with A, we get A ∪ (B ∪ C)' = A ∪ (elements outside B and C)

But in this case, the only elements outside B and C are 8 and 14, which are already in A. So union is still A.

✔ Answer 2: {3,8,9,14,15,16,17}

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3) C ∪ (A - B)

A - B = elements in A but not in B = {14,8,17,15} (since 3,16,9 are in B too)

C = {7,10,18,17,15,9,12,13}

C ∪ (A - B) = {7,10,18,17,15,9,12,13} ∪ {14,8,17,15} = {7,8,9,10,12,13,14,15,17,18}

Note: 17 and 15 are in both, so no duplication.

✔ Answer 3: {7,8,9,10,12,13,14,15,17,18}

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4) A - (B ∩ C)

B ∩ C = {9,12,13}

A = {3,8,9,14,15,16,17}

So A - (B ∩ C) = elements in A but not in B ∩ C = remove 9 from A → {3,8,14,15,16,17}

Note: 12 and 13 are not in A, so we don’t remove them.

✔ Answer 4: {3,8,14,15,16,17}

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5) C - (A ∪ B)

A ∪ B = {3,5,8,9,14,15,16,17,12,13}

C = {7,10,18,17,15,9,12,13}

C - (A ∪ B) = elements in C but not in A ∪ B = remove elements that are in A or B.

From C: 17,15,9,12,13 are in A or B → remove them.

Left: {7,10,18}

✔ Answer 5: {7,10,18}

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6) C ∩ (A ∩ B)'

First, A ∩ B = {3,16,9} (elements in both A and B)

So (A ∩ B)' = U - {3,16,9} = {5,7,8,10,12,13,14,15,17,18}

C = {7,10,18,17,15,9,12,13}

C ∩ (A ∩ B)' = elements in C and not in A ∩ B → remove 9 from C → {7,10,18,17,15,12,13}

✔ Answer 6: {7,10,12,13,15,17,18}

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7) A' ∩ (B - C)'

First, A' = U - A = {5,7,10,12,13,18}

B - C = elements in B but not in C = {3,16,5} (since 9,12,13 are in C)

So (B - C)' = U - {3,16,5} = {7,8,9,10,12,13,14,15,17,18}

Now, A' ∩ (B - C)' = intersection of {5,7,10,12,13,18} and {7,8,9,10,12,13,14,15,17,18}

Common elements: {7,10,12,13,18}

✔ Answer 7: {7,10,12,13,18}

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8) (A ∪ B)' ∪ C

A ∪ B = {3,5,8,9,14,15,16,17,12,13}

(A ∪ B)' = U - (A ∪ B) = {7,10,18}

C = {7,10,18,17,15,9,12,13}

So (A ∪ B)' ∪ C = {7,10,18} ∪ C = C, because {7,10,18} ⊆ C

So answer is C.

✔ Answer 8: {7,9,10,12,13,15,17,18}

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9) (A ∪ C) ∩ B

A ∪ C = {3,8,9,14,15,16,17} ∪ {7,10,18,17,15,9,12,13} = {3,7,8,9,10,12,13,14,15,16,17,18}

B = {3,5,9,12,13,16}

Intersection: common elements → {3,9,12,13,16}

✔ Answer 9: {3,9,12,13,16}

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10) (A ∪ B)' ∪ C'

(A ∪ B)' = {7,10,18} (from earlier)

C' = U - C = {3,5,8,14,16} (since C has 7,9,10,12,13,15,17,18)

So (A ∪ B)' ∪ C' = {7,10,18} ∪ {3,5,8,14,16} = {3,5,7,8,10,14,16,18}

✔ Answer 10: {3,5,7,8,10,14,16,18}

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## ✔ Final Answers:

1) {3,5,7,8,10,12,13,14,15,16,17,18}
2) {3,8,9,14,15,16,17}
3) {7,8,9,10,12,13,14,15,17,18}
4) {3,8,14,15,16,17}
5) {7,10,18}
6) {7,10,12,13,15,17,18}
7) {7,10,12,13,18}
8) {7,9,10,12,13,15,17,18}
9) {3,9,12,13,16}
10) {3,5,7,8,10,14,16,18}

You can write them in any order, as sets are unordered.

Note: Some answers might be written differently (e.g., listing in ascending order), but the elements are correct.

Let me know if you want them boxed or formatted for submission!