The image contains the following modular arithmetic equations to solve:

1. $2x \equiv 6 \pmod{14}$
2. $3x \equiv 9 \pmod{15}$
3. $5x \equiv 20 \pmod{60}$

Let me solve each equation step by step.

1. $2x \equiv 6 \pmod{14}$

First, divide through by the greatest common divisor (GCD) of $2$ and $14$, which is $2$: $x \equiv 3 \pmod{7}$

So the solution is: $x \equiv 3 \pmod{7}$

2. $3x \equiv 9 \pmod{15}$

First, divide through by the GCD of $3$ and $15$, which is $3$: $x \equiv 3 \pmod{5}$

So the solution is: $x \equiv 3 \pmod{5}$

3. $5x \equiv 20 \pmod{60}$

Divide through by the GCD of $5$ and $60$, which is $5$: $x \equiv 4 \pmod{12}$

So the solution is: $x \equiv 4 \pmod{12}$

Final Solutions:

1. $x \equiv 3 \pmod{7}$
2. $x \equiv 3 \pmod{5}$
3. $x \equiv 4 \pmod{12}$

Would you like a more detailed explanation for any step?

Here are five related questions to explore this topic further:

1. How can we solve modular equations where the GCD of coefficients and the modulus is greater than 1?
2. What is the Chinese Remainder Theorem, and how can it be applied to these problems?
3. How can modular arithmetic be used in cryptographic applications?
4. What is the difference between solving modular equations and solving linear congruences?
5. How can modular inverses help when solving equations like $ax \equiv b \pmod{n}$?

Tip: When solving modular equations, always check for common factors (GCD) between the coefficient and the modulus to simplify the problem!