1. **State the problem:** Solve the congruence equation $$2x - 1 \equiv 2 \pmod{5}$$. 2. **Rewrite the equation:** Add 1 to both sides to isolate the term with $x$: $$2x - 1 + 1 \equiv 2 + 1 \pmod{5}$$ which simplifies to $$2x \equiv 3 \pmod{5}$$. 3. **Understand the goal:** We want to find an integer $x$ such that when $2x$ is divided by 5, the remainder is 3. 4. **Find the multiplicative inverse of 2 modulo 5:** Since 5 is prime, the inverse of 2 modulo 5 exists. We look for a number $k$ such that $$2k \equiv 1 \pmod{5}$$. Testing values: - $2 \times 3 = 6 \equiv 1 \pmod{5}$ So, the inverse of 2 modulo 5 is 3. 5. **Multiply both sides of the congruence by 3:** $$3 \times 2x \equiv 3 \times 3 \pmod{5}$$ which simplifies to $$6x \equiv 9 \pmod{5}$$. 6. **Reduce coefficients modulo 5:** $$6x \equiv 9 \pmod{5}$$ becomes $$1x \equiv 4 \pmod{5}$$ since $6 \equiv 1 \pmod{5}$ and $9 \equiv 4 \pmod{5}$. 7. **Final solution:** $$x \equiv 4 \pmod{5}$$ which means $x$ can be any integer of the form $$x = 4 + 5k$$ for any integer $k$. **Answer:** The solution to the congruence is $$x \equiv 4 \pmod{5}$$.