Solve modular arithmetic equations in various moduli.

Problem: Solve the given modular arithmetic equations and find their solutions or truth sets.

#### Equation (a): \( 5x \equiv 8 \pmod{6} \)

1. Simplify the equation modulo 6:
- First, reduce the constant term \( 8 \) modulo 6:
\[
8 \equiv 2 \pmod{6}
\]
- So the equation becomes:
\[
5x \equiv 2 \pmod{6}
\]

2. Find the multiplicative inverse of 5 modulo 6:
- We need an integer \( y \) such that:
\[
5y \equiv 1 \pmod{6}
\]
- Testing values, we find:
\[
5 \cdot 1 = 5 \equiv 5 \pmod{6} \quad (\text{not } 1)
\]
\[
5 \cdot 2 = 10 \equiv 4 \pmod{6} \quad (\text{not } 1)
\]
\[
5 \cdot 3 = 15 \equiv 3 \pmod{6} \quad (\text{not } 1)
\]
\[
5 \cdot 4 = 20 \equiv 2 \pmod{6} \quad (\text{not } 1)
\]
\[
5 \cdot 5 = 25 \equiv 1 \pmod{6} \quad (\text{yes!})
\]
- Therefore, the multiplicative inverse of 5 modulo 6 is \( 5 \).

3. Multiply both sides of the equation by the inverse:
- Multiply both sides of \( 5x \equiv 2 \pmod{6} \) by 5:
\[
5 \cdot 5x \equiv 5 \cdot 2 \pmod{6}
\]
\[
25x \equiv 10 \pmod{6}
\]
- Simplify using \( 25 \equiv 1 \pmod{6} \) and \( 10 \equiv 4 \pmod{6} \):
\[
x \equiv 4 \pmod{6}
\]

4. Solution:
\[
x \equiv 4 \pmod{6}
\]

---

#### Equation (b): \( x - 3 \equiv 4 \pmod{2} \)

1. Simplify the equation modulo 2:
- Add 3 to both sides:
\[
x \equiv 4 + 3 \pmod{2}
\]
\[
x \equiv 7 \pmod{2}
\]
- Reduce 7 modulo 2:
\[
7 \equiv 1 \pmod{2}
\]

2. Solution:
\[
x \equiv 1 \pmod{2}
\]

---

#### Equation (c): \( 6x + 3 \equiv 7 \pmod{4} \)

1. Simplify the equation modulo 4:
- First, reduce each term modulo 4:
\[
6 \equiv 2 \pmod{4}, \quad 3 \equiv 3 \pmod{4}, \quad 7 \equiv 3 \pmod{4}
\]
- The equation becomes:
\[
2x + 3 \equiv 3 \pmod{4}
\]

2. Isolate \( 2x \):
- Subtract 3 from both sides:
\[
2x \equiv 3 - 3 \pmod{4}
\]
\[
2x \equiv 0 \pmod{4}
\]

3. Solve for \( x \):
- Since \( 2x \equiv 0 \pmod{4} \), \( x \) must be a multiple of 2 (because 2 times any odd number is not divisible by 4).
- Let \( x = 2k \) for some integer \( k \). Then:
\[
x \equiv 0 \pmod{2}
\]

4. Solution:
\[
x \equiv 0 \pmod{2}
\]

---

#### Equation (d): \( 5 - 3y \equiv 0 \pmod{8} \)

1. Simplify the equation modulo 8:
- Rearrange the equation:
\[
5 \equiv 3y \pmod{8}
\]
- This is equivalent to:
\[
3y \equiv 5 \pmod{8}
\]

2. Find the multiplicative inverse of 3 modulo 8:
- We need an integer \( z \) such that:
\[
3z \equiv 1 \pmod{8}
\]
- Testing values, we find:
\[
3 \cdot 1 = 3 \equiv 3 \pmod{8} \quad (\text{not } 1)
\]
\[
3 \cdot 2 = 6 \equiv 6 \pmod{8} \quad (\text{not } 1)
\]
\[
3 \cdot 3 = 9 \equiv 1 \pmod{8} \quad (\text{yes!})
\]
- Therefore, the multiplicative inverse of 3 modulo 8 is \( 3 \).

3. Multiply both sides of the equation by the inverse:
- Multiply both sides of \( 3y \equiv 5 \pmod{8} \) by 3:
\[
3 \cdot 3y \equiv 3 \cdot 5 \pmod{8}
\]
\[
9y \equiv 15 \pmod{8}
\]
- Simplify using \( 9 \equiv 1 \pmod{8} \) and \( 15 \equiv 7 \pmod{8} \):
\[
y \equiv 7 \pmod{8}
\]

4. Solution:
\[
y \equiv 7 \pmod{8}
\]

---

#### Equation (e): \( 5a + 3 \equiv 2 \pmod{11} \)

1. Simplify the equation modulo 11:
- Subtract 3 from both sides:
\[
5a \equiv 2 - 3 \pmod{11}
\]
\[
5a \equiv -1 \pmod{11}
\]
- Since \(-1 \equiv 10 \pmod{11}\), the equation becomes:
\[
5a \equiv 10 \pmod{11}
\]

2. Find the multiplicative inverse of 5 modulo 11:
- We need an integer \( w \) such that:
\[
5w \equiv 1 \pmod{11}
\]
- Testing values, we find:
\[
5 \cdot 1 = 5 \equiv 5 \pmod{11} \quad (\text{not } 1)
\]
\[
5 \cdot 2 = 10 \equiv 10 \pmod{11} \quad (\text{not } 1)
\]
\[
5 \cdot 3 = 15 \equiv 4 \pmod{11} \quad (\text{not } 1)
\]
\[
5 \cdot 4 = 20 \equiv 9 \pmod{11} \quad (\text{not } 1)
\]
\[
5 \cdot 5 = 25 \equiv 3 \pmod{11} \quad (\text{not } 1)
\]
\[
5 \cdot 6 = 30 \equiv 8 \pmod{11} \quad (\text{not } 1)
\]
\[
5 \cdot 7 = 35 \equiv 2 \pmod{11} \quad (\text{not } 1)
\]
\[
5 \cdot 8 = 40 \equiv 7 \pmod{11} \quad (\text{not } 1)
\]
\[
5 \cdot 9 = 45 \equiv 1 \pmod{11} \quad (\text{yes!})
\]
- Therefore, the multiplicative inverse of 5 modulo 11 is \( 9 \).

3. Multiply both sides of the equation by the inverse:
- Multiply both sides of \( 5a \equiv 10 \pmod{11} \) by 9:
\[
9 \cdot 5a \equiv 9 \cdot 10 \pmod{11}
\]
\[
45a \equiv 90 \pmod{11}
\]
- Simplify using \( 45 \equiv 1 \pmod{11} \) and \( 90 \equiv 3 \pmod{11} \):
\[
a \equiv 3 \pmod{11}
\]

4. Solution:
\[
a \equiv 3 \pmod{11}
\]

---

Final Answers:

\[
\boxed{
\begin{aligned}
&\text{(a)} \quad x \equiv 4 \pmod{6} \\
&\text{(b)} \quad x \equiv 1 \pmod{2} \\
&\text{(c)} \quad x \equiv 0 \pmod{2} \\
&\text{(d)} \quad y \equiv 7 \pmod{8} \\
&\text{(e)} \quad a \equiv 3 \pmod{11}
\end{aligned}
}
\]