You have a system of linear equations:

1. $y = 6x - 11$
2. $-2x - 3y = -7$

Let's solve it step by step.

Step 1: Substitute $y = 6x - 11$ into the second equation.

Substitute the expression for $y$ from the first equation into the second equation:

$-2x - 3(6x - 11) = -7$

Step 2: Simplify the equation.

Distribute the $-3$ through the parentheses:

$-2x - 18x + 33 = -7$

Combine like terms:

$-20x + 33 = -7$

Step 3: Solve for $x$.

Subtract 33 from both sides:

$-20x = -7 - 33$

$-20x = -40$

Now, divide both sides by $-20$:

$x = \frac{-40}{-20} = 2$

Step 4: Solve for $y$.

Now that we know $x = 2$, substitute this value into the first equation $y = 6x - 11$:

$y = 6(2) - 11 = 12 - 11 = 1$

Solution:

The solution to the system of equations is: $x = 2, \quad y = 1$

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

Here are 5 related questions to explore:

1. How would you solve this system using the substitution method?
2. What would happen if you used the elimination method to solve this system?
3. How do you recognize when a system has no solution or infinitely many solutions?
4. Can you represent the solution of this system graphically?
5. How would you solve a system of linear equations with three variables?

Tip: When solving a system of equations, always check your solution by substituting the values back into the original equations!