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