You have the system of equations:

1. $x + 6y = 11$
2. $2x - 12y = 10$

We can solve this system using the substitution or elimination method. Let's use the elimination method.

Step 1: Eliminate one variable

To eliminate one variable, we can manipulate the equations. Notice that the second equation $2x - 12y = 10$ has a term $2x$, which is just 2 times the first equation's $x$ term.

We will multiply the first equation by 2:

$2(x + 6y) = 2(11)$ $2x + 12y = 22$

So, the system now looks like this:

1. $2x + 12y = 22$
2. $2x - 12y = 10$

Step 2: Subtract the second equation from the first

Now, subtract the second equation from the first:

$(2x + 12y) - (2x - 12y) = 22 - 10$ $2x - 2x + 12y + 12y = 12$ $24y = 12$

Step 3: Solve for $y$

Divide both sides by 24:

$y = \frac{12}{24} = \frac{1}{2}$

Step 4: Substitute $y = \frac{1}{2}$ into one of the original equations

Now, substitute $y = \frac{1}{2}$ into the first equation:

$x + 6y = 11$ $x + 6\left(\frac{1}{2}\right) = 11$ $x + 3 = 11$ $x = 11 - 3$ $x = 8$

Final Answer:

The solution to the system of equations is: $x = 8, \quad y = \frac{1}{2}$

Would you like to see more details or ask any follow-up questions?

Here are some related questions:

1. How would the solution change if the coefficients in the system were different?
2. What would happen if we used substitution instead of elimination?
3. How can we graphically represent this system of equations?
4. What is the significance of the solution in terms of the lines' intersection on a graph?
5. What are the steps to solve a system of three linear equations?

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