What is the solution to the system of equations y = 3x + 2 and 2x - y = -1?

To solve the system of equations:

\[
\begin{align*}
y & = 3x + 2 \quad \text{(Equation 1)} \\
2x - y & = -1 \quad \text{(Equation 2)}
\end{align*}
\]

We can use the substitution method or the elimination method. Here, I'll use the substitution method.

### Step 1: Substitute \( y \) from Equation 1 into Equation 2
Since \( y = 3x + 2 \), substitute this into Equation 2:

\[
2x - (3x + 2) = -1
\]

### Step 2: Simplify the equation
Distribute and combine like terms:

\[
2x - 3x - 2 = -1
\]

\[
-x - 2 = -1
\]

### Step 3: Solve for \( x \)
Add 2 to both sides to isolate \( x \):

\[
-x = 1
\]

Multiply both sides by -1:

\[
x = -1
\]

### Step 4: Solve for \( y \) using the value of \( x \)
Substitute \( x = -1 \) back into Equation 1:

\[
y = 3(-1) + 2
\]

\[
y = -3 + 2 = -1
\]

### Final Answer:
The solution to the system of equations is \( x = -1 \) and \( y = -1 \). Therefore, the solution is the point:

\[
\boxed{(-1, -1)}
\]

Would you like any additional details or explanations?

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Here are 5 follow-up questions you might consider:
1. How would the solution change if the second equation was different?
2. Can you solve the system using the elimination method?
3. What is the graphical interpretation of the solution?
4. How can you verify the solution algebraically?
5. How does the solution relate to the slopes of the lines?

**Tip:** When solving systems of linear equations, it's useful to understand both substitution and elimination methods, as each can be more convenient depending on the specific system.

Explore the solution to the system of equations y = 3x + 2 and 2x - y = -1 with detailed steps using the substitution method. Understand how to solve linear equations and apply algebraic methods effectively. Suitable for Grades 7-9 students and anyone interested in algebraic problem-solving.

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