Solve 3𝑥 + 13𝑦 ≡ 8(𝑚𝑜𝑑 55), 5𝑥 + 21𝑦 ≡ 34(𝑚𝑜𝑑 55)

To solve the given system of congruences, we will use the method of substitution.

Step 1: We start by expressing one variable in terms of the other variable using one of the congruences. Let's solve the first congruence for 𝑥: 3𝑥 + 13𝑦 ≡ 8 (mod 55) 3𝑥 ≡ 8 - 13𝑦 (mod 55)

Step 2: Next, we substitute this expression for 𝑥 in the second congruence: 5(8 - 13𝑦) + 21𝑦 ≡ 34 (mod 55) 40 - 65𝑦 + 21𝑦 ≡ 34 (mod 55) -44𝑦 ≡ -6 (mod 55)

Step 3: We need to find the modular inverse of -44 (mod 55) to solve for 𝑦. The modular inverse of -44 (mod 55) is 24, since -44 * 24 ≡ 1 (mod 55).

Step 4: Multiply both sides of the congruence by the modular inverse: -44𝑦 * 24 ≡ -6 * 24 (mod 55) 𝑦 ≡ -144 (mod 55) 𝑦 ≡ 31 (mod 55)

Step 5: Substitute the value of 𝑦 back into the first congruence to solve for 𝑥: 3𝑥 + 13(31) ≡ 8 (mod 55) 3𝑥 + 403 ≡ 8 (mod 55) 3𝑥 ≡ -395 (mod 55)

Step 6: We need to find the modular inverse of 3 (mod 55) to solve for 𝑥. The modular inverse of 3 (mod 55) is 37, since 3 * 37 ≡ 1 (mod 55).

Step 7: Multiply both sides of the congruence by the modular inverse: 3𝑥 * 37 ≡ -395 * 37 (mod 55) 𝑥 ≡ -14615 (mod 55) 𝑥 ≡ 20 (mod 55)

Therefore, the solution to the system of congruences is 𝑥 ≡ 20 (mod 55) and 𝑦 ≡ 31 (mod 55).