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The problem asks you to solve the linear congruence:

\[
74x \equiv 7 \pmod{69}
\]

### Step-by-Step Solution:

1. **Simplify the equation mod 69**:
   \[
   74x \equiv 7 \pmod{69}
   \]
   Notice that \( 74 \equiv 5 \pmod{69} \), since \( 74 - 69 = 5 \). So the equation simplifies to:
   \[
   5x \equiv 7 \pmod{69}
   \]

2. **Find the multiplicative inverse of 5 mod 69**:
   To solve \( 5x \equiv 7 \pmod{69} \), we need to find the inverse of 5 modulo 69. This can be done using the extended Euclidean algorithm.

   Applying the algorithm:
   \[
   69 = 13 \times 5 + 4
   \]
   \[
   5 = 1 \times 4 + 1
   \]
   \[
   4 = 4 \times 1 + 0
   \]
   Working backward, we find:
   \[
   1 = 5 - 1 \times (69 - 13 \times 5) = 14 \times 5 - 1 \times 69
   \]
   So, the inverse of 5 mod 69 is \( 14 \), since \( 14 \times 5 \equiv 1 \pmod{69} \).

3. **Solve for x**:
   Multiply both sides of the simplified congruence \( 5x \equiv 7 \pmod{69} \) by the inverse of 5, which is 14:
   \[
   x \equiv 14 \times 7 \pmod{69}
   \]
   Calculate:
   \[
   14 \times 7 = 98
   \]
   Now reduce 98 modulo 69:
   \[
   98 \equiv 29 \pmod{69}
   \]

So, the solution is:
\[
x \equiv 29 \pmod{69}
\]

### Final Answer:
\[
x = 29 \pmod{69}
\]

Let me know if you need more details or explanations!

Here are 5 related questions to deepen your understanding:

1. What is the general method to find the inverse of a number in modular arithmetic?
2. How would the solution change if the modulus were different, say mod 70?
3. Can all linear congruences be solved using the method outlined?
4. What if the coefficient of \(x\) and the modulus are not coprime? How do you handle that case?
5. How does the extended Euclidean algorithm work, and why is it useful in modular arithmetic?

**Tip**: When solving linear congruences, always check if the greatest common divisor (GCD) of the coefficient of \(x\) and the modulus is 1, as this ensures that an inverse exists.

Solve the linear congruence 74x ≡ 7 (mod 69).

Learn how to solve the linear congruence 74x ≡ 7 (mod 69) with a detailed step-by-step solution, applying the extended Euclidean algorithm to find the modular inverse of 5 mod 69. This problem explores key modular arithmetic concepts suitable for high school students.

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