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.