Equal sets and Subsets

The intersection contains elements common to both sets. Since \(A\subseteq B\), every element of A is also an element of B. Thus all elements of A are common to A and B, while B may contain additional elements that are not in A. Therefore, \(A\cap B=A\). The answer does not depend on the particular elements, only on the subset relationship.

People with only a mobile phone number 92 − 51 = 41, because those with both devices must be removed. People with only a laptop number 78 − 51 = 27. Exactly one device means only mobile or only laptop, so the required number is 41 + 27 = 68. Therefore, option A is correct.

Since A⊆B and A⊆C, every element of A belongs to all three sets. Conversely, an element in A∩B∩C must certainly belong to A. Thus the intersection cannot contain anything outside A and contains every element of A, so A∩B∩C=A. B and C may contain additional elements, while C−A contains elements specifically excluded from A.

Y − X contains elements that are in Y but not in X. The elements of Y are 3, 7, and 9. Since 3 and 7 also belong to X, they must be removed. The element 9 is not in X, so it remains. Consequently, Y − X = {9}. The set {3, 7} is Y ∩ X, while {1, 5} belongs to X − Y, not Y − X.

The relation \(A\subseteq B\) means that every element of \(A\) is already contained in \(B\). The union collects all elements from both sets, but adding the elements of \(A\) contributes nothing new because they are already in \(B\). Consequently, \(A\cup B=B\). In contrast, \(A\cap B=A\) and \(A\setminus B=\varnothing\) under this condition.

Because \(A\subseteq B\), every element of \(A\) is also an element of \(B\). The intersection contains elements common to both sets, so it contains every element of \(A\) and no element outside \(A\) is required. Therefore, \(A\cap B=A\). Set \(B\) may have additional elements, the union is generally larger, and \(B\setminus A\) contains elements of \(B\) excluded from \(A\).

Taking the union of a set with itself does not introduce any new elements. Every element of A is already in the first copy and the second copy, so A ∪ A = A. This is called the idempotent law. In contrast, A − A = ∅ and A ∩ ∅ = ∅, so those expressions are not generally equal to A.

The equality \(A\cap B=A\) means that intersecting A with B removes nothing from A. Consequently, every element of A must already be in B, so \(A\subseteq B\). The reverse relation \(B\subseteq A\) is not required; for example, A may be a proper subset of B. The complement and empty-union statements also do not follow. Therefore option A is the uniquely necessary conclusion.

The union \(A\cup B\) contains all elements of A and all elements of B. If the union is equal to A, then B cannot contain any element outside A; otherwise that element would appear in the union and make it larger than A. Hence every element of B belongs to A, which means \(B\subseteq A\). The other choices are not necessary and may contradict this containment.

The difference A \ B consists of elements that belong to A but do not belong to B. If this difference is empty, there is no element of A outside B. Therefore, every element of A is also an element of B, which is exactly the definition of A ⊆ B. The other statements do not necessarily follow; for example, A and B may overlap without being equal.

Factor the equation as a² - b² = (a - b)(a + b) = 0. Since a and b are both positive elements of {1,2,3,4,5}, the equation a + b = 0 is impossible. Therefore, a - b = 0, so a = b. The valid ordered pairs are (1,1), (2,2), (3,3), (4,4), and (5,5). There are exactly five such pairs, so option B is correct.

Two ordered pairs are equal if and only if their corresponding components are equal. Therefore, compare the first components and the second components separately: 2x - 1 = 5 and y + 3 = 10. The first equation gives 2x = 6, so x = 3. The second gives y = 7. Hence (x, y) = (3, 7), making option B correct. The order of components must be preserved while comparing ordered pairs.

Check the condition separately for each x in A. If x = 1, even the largest sum is 1 + 8 = 9, so no pair qualifies. If x = 3, only y = 8 gives a sum greater than 9, contributing one pair. If x = 5, y = 6 and y = 8 give sums 11 and 13, contributing two pairs. The total is 0 + 1 + 2 = 3, so option B is correct.

Count the allowable values of \(y\) for each value of \(x\). If \(x=1\), then \(y\ge3\), so \(y=3,4,5\): three pairs. If \(x=2\), then \(y\ge4\), so \(y=4,5\): two pairs. If \(x=3\), then \(y\ge5\), so only \(y=5\): one pair. Hence the total is \(3+2+1=6\), making option A correct.