Answer
1. **\(A \subseteq B\)** is **false**.
2. **\(B \subseteq A\)** is **true**.
3. **\(B \subseteq C\)** is **true** (indeed \(B=C\)).
Work Step by Step
First, let us restate each set in a more convenient (modular) form:
- \(A = \{\,5a + 2 : a \in \mathbb{Z}\}\).
Equivalently, \(A\) is the set of all integers congruent to \(2\pmod{5}\).
- \(B = \{\,10b - 3 : b \in \mathbb{Z}\}\).
Note that \(10b - 3 \equiv -3 \equiv 7 \pmod{10}\).
Thus \(B\) is the set of all integers congruent to \(7\pmod{10}\).
- \(C = \{\,10c + 7 : c \in \mathbb{Z}\}\).
Clearly \(C\) is also the set of all integers congruent to \(7\pmod{10}\).
From the last two bullets, one immediately sees \(B\) and \(C\) are **the same set** (both are “all integers that are \(7\pmod{10}\)”). We now check each of the three statements.
---
## (a) \(A \subseteq B\)?
To have \(A \subseteq B\), *every* element of \(A\) would need to lie in \(B\). But:
- An element of \(A\) is any integer \(x\equiv 2\pmod{5}\).
- An element of \(B\) is any integer \(y\equiv 7\pmod{10}\).
A quick counterexample:
\[
x = 2 \;\;(\text{which is }5\cdot0 + 2)\quad\text{is in }A,
\]
but \(2 \not\equiv 7\pmod{10}\). Hence \(2\notin B\).
Thus there exists an element in \(A\) that is *not* in \(B\), so
\[
\boxed{A \subseteq B\text{ is false.}}
\]
---
## (b) \(B \subseteq A\)?
To have \(B \subseteq A\), *every* element of \(B\) must lie in \(A\). Check:
- If \(x \in B\), then \(x \equiv 7 \pmod{10}\).
- Reducing modulo \(5\), we get \(x \equiv 7 \equiv 2 \pmod{5}\).
Hence any \(x\equiv 7\pmod{10}\) also satisfies \(x \equiv 2\pmod{5}\), which means \(x\in A\). Therefore
\[
\boxed{B \subseteq A\text{ is true.}}
\]
---
## (c) \(B \subseteq C\)?
Recall:
- \(B\) is all integers congruent to \(7\pmod{10}\).
- \(C\) is also all integers congruent to \(7\pmod{10}\).
Hence \(B\) *equals* \(C\). In particular, \(B \subseteq C\) trivially holds (and so does \(C \subseteq B\)). Thus
\[
\boxed{B \subseteq C\text{ is true (in fact }B=C\text{).}}
\]
