Thank you for providing the equations! The system of linear equations is:

1. $6x + 5y + 4 = 0$
2. $3x - 4y = 11$

Let's solve this system of equations using either the substitution method or the elimination method. I'll proceed with the elimination method for clarity.

Step 1: Eliminate one of the variables

First, we’ll rewrite the equations in standard form:

$6x + 5y = -4 \quad \text{(Equation 1)}$ $3x - 4y = 11 \quad \text{(Equation 2)}$

Step 2: Multiply the equations to align coefficients for elimination

To eliminate $x$, we can multiply the second equation by 2:

$2(3x - 4y) = 2(11)$ $6x - 8y = 22 \quad \text{(New Equation 2)}$

Now, we have the system: $6x + 5y = -4 \quad \text{(Equation 1)}$ $6x - 8y = 22 \quad \text{(New Equation 2)}$

Step 3: Subtract the equations

Now subtract Equation 1 from New Equation 2:

$(6x - 8y) - (6x + 5y) = 22 - (-4)$ Simplifying the left-hand side: $6x - 8y - 6x - 5y = 26$ $-13y = 26$

Step 4: Solve for $y$

Now solve for $y$: $y = \frac{26}{-13} = -2$

Step 5: Substitute $y = -2$ back into one of the original equations

Let’s substitute $y = -2$ into Equation 2: $3x - 4(-2) = 11$ $3x + 8 = 11$ Subtract 8 from both sides: $3x = 3$ Divide by 3: $x = 1$

Final Solution:

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

Would you like a detailed explanation of any step, or do you have any other questions related to this problem?

Here are five related questions to further your understanding:

1. How would you solve this system using the substitution method?
2. Can you express the solution graphically?
3. What happens when the system of equations has no solutions?
4. How can you verify the solution for this system of equations?
5. Can the method of elimination be applied to systems with three variables?

Tip: Always check your solution by substituting the values of $x$ and $y$ back into the original equations to verify accuracy.