Solution. Two inclusions, as always.

\(\subset \). Let \(x\in A\cap (B\cup C)\). Then \(x\in A\), and \(x\in B\cup C\), so \(x\in B\) or \(x\in C\).

If \(x\in B\), then since also \(x\in A\) we have \(x\in A\cap B\).

If \(x\in C\), then likewise \(x\in A\cap C\).

Either way \(x\in (A\cap B)\cup (A\cap C)\).

\(\supset \). Let \(x\in (A\cap B)\cup (A\cap C)\), so \(x\in A\cap B\) or \(x\in A\cap C\). In the first case \(x\in A\) and \(x\in B\); in the second \(x\in A\) and \(x\in C\). In both cases \(x\in A\), and in both cases \(x\) lies in \(B\) or in \(C\), so \(x\in B\cup C\). Hence \(x\in A\cap (B\cup C)\).

Both inclusions hold, so the sets are equal.

Solution. (a). By definition of difference, \(E-A=\{x: x\in E \text { and } x\notin A\}\), which is precisely the definition of \(A'\). The two are the same set by definition rather than by argument.

(b). \(x\in A-B\) means \(x\in A\) and \(x\notin B\). Since \(x\notin B\) is the same as \(x\in B'\), this says \(x\in A\) and \(x\in B'\), that is, \(x\in A\cap B'\). Every step reverses, so the two sets have the same elements.

(c). If \(x\in A-B\) then in particular \(x\in A\). Every element of \(A-B\) is therefore an element of \(A\), which is what \(A-B\subset A\) means. Note that removing elements from a set can only shrink it, never enlarge it.

Solution. Let \(M\) and \(P\) be the sets taking Mathematics and Physics, with \(E\) the class, so \(n(E)=40\). Since \(7\) take neither, the number taking at least one is \[n(M\cup P)=40-7=33 .\] Counting \(n(M)+n(P)\) counts anyone taking both subjects twice, so \[n(M\cup P)=n(M)+n(P)-n(M\cap P),\] \[33=25+18-n(M\cap P)=43-n(M\cap P),\] giving \(n(M\cap P)=10\). Ten learners take both.

As a check, the four groups are: both \(10\), Mathematics only \(25-10=15\), Physics only \(18-10=8\), neither \(7\). These total \(10+15+8+7=40\), the whole class.

Solution.

(a).

\(A=\{x: -10<x\leq 10,\ x\in \mathbb {R}\}\)

(b).

\(B=\{x: 4\leq x\leq 9,\ x\in \mathbb {R}\}\)

(c).

\(C=\{x: -1<x<5,\ x\in \mathbb {R}\}\)

(d).

\(D=\{x: 0\leq x<-2,\ x\in \mathbb {R}\}=\emptyset \)

Part (d) is worth pausing on. No real number is both at least \(0\) and less than \(-2\), so the condition is never satisfied and \(D\) is empty. Interval notation always lists the smaller endpoint first, and \([0,-2)\) breaks that rule — so either it is a slip for \([-2,0)\), or it denotes the empty set. Read literally, it is the latter.

Solution. First write \(C\) as an interval: \(C=[4,\infty )\), so \(C'=(-\infty ,4)\). Also \(B'=(-\infty ,0]\cup [4,\infty )\).

(a). \(A-C\) removes from \((-1,6)\) everything from \(4\) upwards: \[A-C=(-1,4).\]

(b). \(B'\cap C\) is the part of \(B'\) lying in \([4,\infty )\), which is \([4,\infty )\) itself. Its complement is \[\left (B'\cap C\right )'=(-\infty ,4).\]

(c). \(C-A\) removes \((-1,6)\) from \([4,\infty )\), leaving \([6,\infty )\). But \(B=(0,4)\) contains nothing from \(6\) upwards, so \[(C-A)\cap B=\emptyset .\] Sketching the three sets on one number line makes all three answers readable without algebra, which is what the question intends.

Solution. In interval notation \(A=(-\infty ,2]\) and \(B=(-6,\infty )\), so \[A'=(2,\infty ).\] Now \(A'\cap B\) needs points in both \((2,\infty )\) and \((-6,\infty )\); since the first is contained in the second, \(A'\cap B=(2,\infty )\). Taking the complement, \[\left (A'\cap B\right )'=(-\infty ,2]=A .\] On the number line this is everything from \(2\) leftwards, with a filled circle at \(2\) to show it is included.

Solution. Work outwards from the innermost bracket. \[B'=[-10,-5]\cup [3,10],\] \[B'\cup C=[-10,-5]\cup [-1,8]\cup [3,10]=[-10,-5]\cup [-1,10],\] the last two overlapping and merging. Its complement within \(U\) is what is left of \([-10,10]\): \[\left (B'\cup C\right )'=(-5,-1).\] Finally, \(A=[-2,6)\) and \((-5,-1)\) overlap, so their union is a single interval: \[A\cup \left (B'\cup C\right )'=(-5,6).\]