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Let's solve each of these linear congruence problems step-by-step.

### Problem 1
**Equation**: \( 7x \equiv 5 \pmod{11} \)

To solve for \( x \), we need to find the modular inverse of 7 modulo 11, which is a number \( y \) such that \( 7y \equiv 1 \pmod{11} \).

1. The inverse of 7 modulo 11 is 8 (since \( 7 \times 8 \equiv 1 \pmod{11} \)).
2. Multiply both sides by 8:
   \[
   x \equiv 5 \times 8 \pmod{11}
   \]
3. Simplify:
   \[
   x \equiv 40 \pmod{11} \equiv 7 \pmod{11}
   \]
**Solution**: \( x \equiv 7 \pmod{11} \)

---

### Problem 2
**Equation**: \( 9x \equiv 4 \pmod{13} \)

To solve for \( x \), we need the modular inverse of 9 modulo 13.

1. The inverse of 9 modulo 13 is 3 (since \( 9 \times 3 \equiv 1 \pmod{13} \)).
2. Multiply both sides by 3:
   \[
   x \equiv 4 \times 3 \pmod{13}
   \]
3 \ solution

Solve each linear congruence problem:
1. Solve for x in the congruence 7x ≡ 5 (mod 11).
2. Solve for x in the congruence 9x ≡ 4 (mod 13).
3. Solve for x in the congruence 8x ≡ 12 (mod 20).
4. Determine if a solution exists for 10x ≡ 7 (mod 15), and solve for x if it exists.
5. Find the modular inverse of 3 modulo 11, then use it to solve 3x ≡ 8 (mod 11).
6. Solve for x in the congruence 6x ≡ 9 (mod 15).

Learn how to solve linear congruences like 7x ≡ 5 (mod 11) and find modular inverses in modular arithmetic. This guide provides step-by-step solutions to six example problems, covering key concepts in modular arithmetic and linear congruences. Suitable for high school students in Grades 10-12.

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