We are tasked with solving the congruence:

\[
64x + 11 \equiv 43 \pmod{84}
\]

### Step-by-step solution:

1. **Simplify the equation**:
   First, subtract 11 from both sides to isolate the term with \(x\):
   \[
   64x \equiv 43 - 11 \pmod{84}
   \]
   \[
   64x \equiv 32 \pmod{84}
   \]

2. **Reduce modulo 84**:
   We now need to simplify the coefficient of \(x\) and the right-hand side. The greatest common divisor (GCD) of 64 and 84 is 4. We divide the entire equation by 4:
   \[
   \frac{64x}{4} \equiv \frac{32}{4} \pmod{\frac{84}{4}}
   \]
   \[
   16x \equiv 8 \pmod{21}
   \]

3. **Solve for \(x\)**:
   We now solve the equation \(16x \equiv 8 \pmod{21}\). To do this, we need to find the modular inverse of 16 modulo 21. We can use the extended Euclidean algorithm to do this.

4. **Use the extended Euclidean algorithm**:
   We need to find the inverse of 16 modulo 21 by solving:
   \[
   16y \equiv 1 \pmod{21}
   \]
   Applying the extended Euclidean algorithm, we find that the inverse of 16 modulo 21 is 4.

5. **Multiply both sides by the inverse**:
   Multiply both sides of \(16x \equiv 8 \pmod{21}\) by 4:
   \[
   x \equiv 8 \times 4 \pmod{21}
   \]
   \[
   x \equiv 32 \pmod{21}
   \]
   Since \(32 \equiv 11 \pmod{21}\), we have:
   \[
   x \equiv 11 \pmod{21}
   \]

Thus, the solution to the congruence is:
\[
x \equiv 11 \pmod{21}
\]

Would you like any further details or clarifications?

Here are some related questions for practice:
1. Solve \(45x + 7 \equiv 25 \pmod{60}\).
2. Solve \(27x \equiv 18 \pmod{81}\).
3. Solve \(35x + 20 \equiv 15 \pmod{50}\).
4. Find the inverse of 17 modulo 30.
5. Solve \(13x \equiv 1 \pmod{19}\).

**Tip:** When solving modular equations, if the GCD of the coefficient and the modulus is not 1, make sure to simplify the equation by dividing by the GCD if possible.

This page provides a detailed step-by-step solution to the congruence equation 64x + 11 ≡ 43 (mod 84). Learn how to simplify the equation, apply the extended Euclidean algorithm to find modular inverses, and solve for x. Suitable for students in Grades 10-12.

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