To solve the linear congruence:

$2x \equiv 5 \ (\text{mod} \ 7)$

we need to find an integer $x$ such that when multiplied by 2, the result is congruent to 5 modulo 7. The general approach involves finding the modular inverse of 2 modulo 7 and multiplying both sides of the congruence by this inverse.

Step 1: Find the modular inverse of 2 modulo 7

The modular inverse of $2$ modulo $7$ is a number $y$ such that:

$2y \equiv 1 \ (\text{mod} \ 7)$

We can check values of $y$ from 1 to 6 to find which satisfies the equation:

• $2 \times 1 = 2$
• $2 \times 2 = 4$
• $2 \times 3 = 6$
• $2 \times 4 = 8 \equiv 1 \ (\text{mod} \ 7)$

Thus, the modular inverse of 2 modulo 7 is $4$.

Step 2: Multiply both sides of the congruence by the modular inverse

Now, multiply both sides of the original congruence $2x \equiv 5 \ (\text{mod} \ 7)$ by 4:

$4 \cdot (2x) \equiv 4 \cdot 5 \ (\text{mod} \ 7)$

This simplifies to:

$8x \equiv 20 \ (\text{mod} \ 7)$

Since $8 \equiv 1 \ (\text{mod} \ 7)$, the equation reduces to:

$x \equiv 20 \ (\text{mod} \ 7)$

Step 3: Simplify the right-hand side modulo 7

Now, reduce $20 \mod 7$:

$20 \div 7 = 2 \ \text{(quotient)}, \quad 20 - 2 \times 7 = 6$

Thus, $20 \equiv 6 \ (\text{mod} \ 7)$.

Final Solution:

Therefore, the solution to the congruence is:

$x \equiv 6 \ (\text{mod} \ 7)$

This means $x = 6$ is the smallest solution.

Would you like a more detailed explanation, or do you have any questions?

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Here are 5 related questions:

1. What are the steps to finding a modular inverse in general?
2. How do we verify if a number has a modular inverse in a given modulus?
3. Can a linear congruence always be solved? Under what conditions?
4. What are some practical applications of solving linear congruences?
5. How does solving this congruence relate to solving systems of modular equations?

Tip: When solving congruences, always check if the coefficients are relatively prime to the modulus to ensure a solution exists.