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To solve the given system of equations using the substitution method, we have:

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

### Step 1: Solve for \( x \) in terms of \( y \) using the first equation
From the first equation:
\[
x - 3 = 3y
\]
we can rearrange this to solve for \( x \):
\[
x = 3y + 3
\]

### Step 2: Substitute \( x \) from the first equation into the second equation
Now, substitute \( x = 3y + 3 \) into the second equation:
\[
(3y + 3) y = 6
\]

Expanding and simplifying:
\[
3y^2 + 3y = 6
\]
\[
3y^2 + 3y - 6 = 0
\]

### Step 3: Simplify the quadratic equation
Divide through by 3 to make it simpler:
\[
y^2 + y - 2 = 0
\]

### Step 4: Solve the quadratic equation for \( y \)
The equation \( y^2 + y - 2 = 0 \) can be factored as:
\[
(y + 2)(y - 1) = 0
\]

This gives two possible values for \( y \):
\[
y = -2 \quad \text{or} \quad y = 1
\]

### Step 5: Find the corresponding \( x \) values
For each \( y \), substitute back into \( x = 3y + 3 \) to find \( x \).

1. If \( y = -2 \):
   \[
   x = 3(-2) + 3 = -6 + 3 = -3
   \]
   So, one solution is \( (x, y) = (-3, -2) \).

2. If \( y = 1 \):
   \[
   x = 3(1) + 3 = 3 + 3 = 6
   \]
   So, the other solution is \( (x, y) = (6, 1) \).

### Final Answer
The solutions to the system are:
\[
(x, y) = (-3, -2) \quad \text{and} \quad (x, y) = (6, 1)
\]

Would you like more details or have any questions?

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Here are some related questions to expand your understanding:
1. How would the solution process change if we used the elimination method instead?
2. Can systems of linear equations always be solved by substitution?
3. What does it mean geometrically when a system has two solutions, as we found here?
4. How would the solution change if the second equation was \( xy = 9 \) instead of \( xy = 6 \)?
5. Can this system of equations be solved by graphing? If so, what would the intersection points represent?

**Tip:** When using substitution, always isolate one variable in terms of the other from the simpler equation to make the process smoother.

Решите систему уравнений методом подстановки: x - 3 = 3y и xy = 6.

Learn how to solve the system of equations x - 3 = 3y and xy = 6 using the substitution method. This step-by-step solution covers key algebraic concepts and the quadratic formula, ideal for students in Grades 9-11.

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