Word Count : 542

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We have a system of congruences: 1. \(x \equiv 2 \pmod{17}\) 2. \(3x \equiv 4 \pmod{19}\) 3. \(x \equiv 7 \pmod{23}\) We will solve this system using the method of successive substitution or the Chinese Remainder Theorem. ### Step 1: Solve the first and second congruences We start with: \[ x \equiv 2 \pmod{17} \] So, we can write \(x = 17k + 2\) for some integer \(k\). Now, substitute this into the second congruence: \[ 3x \equiv 4 \pmod{19} \] Substituting \(x = 17k + 2\), we get: \[ 3(17k + 2) \equiv 4 \pmod{19} \] Simplifying: \[ 51k + 6 \equiv 4 \pmod{19} \] Since \(51 \equiv 13 \pmod{19}\) and \(6 \equiv 6 \pmod{19}\), this becomes: \[ 13k + 6 \equiv 4 \pmod{19} \] Subtract 6 from both sides: \[ 13k \equiv -2 __________ _______ ______ ____ ___ ____.
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