We will use the method of substitution to solve the given system of equations:

\[
2x + 3y = -7 \quad \text{(Equation 1)}
\]
\[
2x - y = 1 \quad \text{(Equation 2)}
\]

### Step 1: Solve one equation for one variable
Let's solve **Equation 2** for \( y \).

\[
2x - y = 1
\]

Rearrange it to solve for \( y \):

\[
y = 2x - 1 \quad \text{(Equation 3)}
\]

### Step 2: Substitute the expression for \( y \) into **Equation 1**

Now, substitute the expression \( y = 2x - 1 \) from **Equation 3** into **Equation 1**:

\[
2x + 3(2x - 1) = -7
\]

### Step 3: Solve for \( x \)

Expand the equation:

\[
2x + 6x - 3 = -7
\]

Combine like terms:

\[
8x - 3 = -7
\]

Add 3 to both sides:

\[
8x = -4
\]

Now, divide by 8:

\[
x = -\frac{1}{2}
\]

### Step 4: Substitute \( x = -\frac{1}{2} \) back into **Equation 3**

Now that we have \( x = -\frac{1}{2} \), substitute it back into **Equation 3** to solve for \( y \):

\[
y = 2(-\frac{1}{2}) - 1
\]

Simplify:

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

### Final Solution:

The solution to the system of equations is:

\[
x = -\frac{1}{2}, \quad y = -2
\]

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

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Here are 5 related questions to expand on this topic:

1. What is the substitution method in solving systems of equations?
2. Can substitution be used for systems of nonlinear equations?
3. How does the elimination method compare to substitution for solving systems?
4. What are some applications of solving systems of equations in real life?
5. How would this method change if we had three variables instead of two?

**Tip:** Always check your solution by plugging the values of \( x \) and \( y \) back into the original equations!

Learn how to solve the system of linear equations 2x+3y = -7 and 2x-y = 1 using the substitution method. This detailed step-by-step solution includes algebraic concepts and substitution techniques, ideal for students in Grades 8-10.

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