We have the system of linear equations:

\[
-7x - 3y = -20 \quad \text{(1)}
\]
\[
3x + 6y = -15 \quad \text{(2)}
\]

We will solve this system using either substitution or elimination. Let's use the **elimination method** to eliminate one variable.

### Step 1: Multiply the first equation to align terms
First, let's multiply equation (1) by 2 to align the coefficients of \(y\):

\[
2(-7x - 3y) = 2(-20)
\]
\[
-14x - 6y = -40 \quad \text{(3)}
\]

Now we have:
\[
-14x - 6y = -40 \quad \text{(3)}
\]
\[
3x + 6y = -15 \quad \text{(2)}
\]

### Step 2: Add equations (3) and (2)
Now, add the two equations to eliminate \(y\):

\[
(-14x - 6y) + (3x + 6y) = -40 + (-15)
\]
\[
-11x = -55
\]

### Step 3: Solve for \(x\)
\[
x = \frac{-55}{-11} = 5
\]

### Step 4: Substitute \(x = 5\) into one of the original equations
Let's substitute \(x = 5\) into equation (2):

\[
3(5) + 6y = -15
\]
\[
15 + 6y = -15
\]
\[
6y = -15 - 15
\]
\[
6y = -30
\]
\[
y = \frac{-30}{6} = -5
\]

### Final Solution:
The solution to the system is:
\[
x = 5, \quad y = -5
\]

Let me know if you need further details!

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Here are 5 related questions:
1. What is the substitution method for solving systems of equations?
2. How do you graphically interpret the solution of a system of linear equations?
3. Can you solve a system of equations using matrices?
4. What happens if a system of equations has no solution?
5. How do you solve systems of equations with more than two variables?

**Tip:** Always double-check your solutions by plugging them back into the original equations to verify their correctness.

Learn how to solve the system of linear equations -7x - 3y = -20 and 3x + 6y = -15 using the elimination method. This step-by-step solution explains the elimination process to find x = 5 and y = -5, highlighting key algebraic concepts. Suitable for students in Grades 8-10.

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