We can solve the system of linear equations:

\[
1. \quad 2x + 3y = 10
\]
\[
2. \quad -2x - y = -2
\]

### Step 1: Add the two equations

First, observe that adding the two equations will eliminate \( x \). Adding equation (1) and equation (2) gives:

\[
(2x + 3y) + (-2x - y) = 10 + (-2)
\]

Simplifying:

\[
2x - 2x + 3y - y = 10 - 2
\]
\[
0x + 2y = 8
\]
\[
2y = 8
\]

### Step 2: Solve for \( y \)

Divide both sides by 2:

\[
y = \frac{8}{2} = 4
\]

### Step 3: Substitute \( y = 4 \) back into one of the original equations

Substitute \( y = 4 \) into the first equation \( 2x + 3y = 10 \):

\[
2x + 3(4) = 10
\]
\[
2x + 12 = 10
\]
\[
2x = 10 - 12
\]
\[
2x = -2
\]

### Step 4: Solve for \( x \)

Divide by 2:

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

### Solution:
The solution to the system of equations is:

\[
x = -1, \quad y = 4
\]

Would you like more details or have any questions?

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Here are 5 related questions you might want to consider:

1. How can you verify that \( x = -1 \) and \( y = 4 \) is the correct solution?
2. How do substitution and elimination methods compare when solving linear systems?
3. What are the conditions for a system of equations to have no solution or infinitely many solutions?
4. How would this system look graphically and what does the solution represent in terms of the graph?
5. How can you solve this system using matrices or determinants?

**Tip:** Always double-check your solution by substituting the values back into both original equations.

Explore the step-by-step solution to the system of equations 2x + 3y = 10 and -2x - y = -2. This detailed guide covers essential algebraic concepts and techniques suitable for students in Grades 7-9.

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