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Let's solve the two systems of equations as requested.

### Задание 2
Solve the system using the substitution method:

\[
\begin{cases}
x - 3 = 3y \\
xy = 6
\end{cases}
\]

1. From the first equation, express \( x \) in terms of \( y \):
   \[
   x = 3y + 3
   \]

2. Substitute \( x = 3y + 3 \) into the second equation:
   \[
   (3y + 3)y = 6
   \]

3. Expand and simplify:
   \[
   3y^2 + 3y = 6
   \]
   \[
   3y^2 + 3y - 6 = 0
   \]
   \[
   y^2 + y - 2 = 0
   \]

4. Factor the quadratic equation:
   \[
   (y + 2)(y - 1) = 0
   \]
   So, \( y = -2 \) or \( y = 1 \).

5. Substitute each value of \( y \) back into \( x = 3y + 3 \) to find \( x \):
   - For \( y = 1 \):
     \[
     x = 3(1) + 3 = 6
     \]
     Solution: \( (x, y) = (6, 1) \)
   
   - For \( y = -2 \):
     \[
     x = 3(-2) + 3 = -6 + 3 = -3
     \]
     Solution: \( (x, y) = (-3, -2) \)

**Answer for Задание 2**: \((6, 1)\) and \((-3, -2)\), which matches the given answer.

---

### Задание 3
Solve the system using the method of elimination:

\[
\begin{cases}
4y^2 - 2xy = 12 \\
y^2 + 8xy = 20
\end{cases}
\]

1. Multiply the second equation by 4 to align the terms:
   \[
   4y^2 + 32xy = 80
   \]

2. Now we have:
   \[
   \begin{cases}
   4y^2 - 2xy = 12 \\
   4y^2 + 32xy = 80
   \end{cases}
   \]

3. Subtract the first equation from the second:
   \[
   (4y^2 + 32xy) - (4y^2 - 2xy) = 80 - 12
   \]
   \[
   34xy = 68
   \]
   \[
   xy = 2
   \]

4. Substitute \( xy = 2 \) back into the first equation:
   \[
   4y^2 - 2(2) = 12
   \]
   \[
   4y^2 - 4 = 12
   \]
   \[
   4y^2 = 16
   \]
   \[
   y^2 = 4
   \]
   \[
   y = \pm 2
   \]

5. Substitute each value of \( y \) into \( xy = 2 \) to find \( x \):
   - For \( y = 2 \):
     \[
     x \cdot 2 = 2 \Rightarrow x = 1
     \]
     Solution: \( (x, y) = (1, 2) \)
   
   - For \( y = -2 \):
     \[
     x \cdot (-2) = 2 \Rightarrow x = -1
     \]
     Solution: \( (x, y) = (-1, -2) \)

**Answer for Задание 3**: \((1, 2)\) and \((-1, -2)\), which matches the given answer.

---

Would you like more details on any step, or do you have any specific questions?

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### Related Questions
1. How can substitution help solve non-linear systems of equations?
2. What are common methods for solving quadratic equations?
3. How can the elimination method simplify systems with similar terms?
4. What other techniques can be used for systems of equations beyond substitution and elimination?
5. When is factoring effective in solving equations?

### Tip
Always double-check each solution by plugging the values back into the original equations to verify accuracy.

Задание 2: Решите систему уравнений методом подстановки. 
\( x - 3 = 3y \\
xy = 6 \)
Задание 3: Решите систему уравнений методом алгебраического сложения. 
\( 4y^2 - 2xy = 12 \\
y^2 + 8xy = 20 \)

Learn how to solve non-linear systems of equations using substitution and elimination methods. This example covers key algebraic techniques like factoring and eliminating variables, with step-by-step solutions. Suitable for students in Grades 9-11.

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