The system of equations you've written appears to be:

1. $4x + 3y = 8$
2. $4x + 6y = 20$

To solve this system, we can use either the substitution method or the elimination method. Let's use the elimination method since both equations have the same $4x$ term.

Step-by-step solution:

1. Subtract the first equation from the second equation to eliminate $x$:

$(4x + 6y) - (4x + 3y) = 20 - 8$ $4x + 6y - 4x - 3y = 12$ $3y = 12$

2. Solve for $y$:

$y = \frac{12}{3} = 4$

3. Substitute $y = 4$ back into the first equation to solve for $x$:

$4x + 3(4) = 8$ $4x + 12 = 8$ $4x = 8 - 12$ $4x = -4$ $x = \frac{-4}{4} = -1$

Solution:

$x = -1, \quad y = 4$

Do you need more details or have any questions?

Here are 5 related questions for practice:

1. Solve the system: $3x + 2y = 5$ and $6x + 4y = 10$.
2. Solve the system: $2x + y = 3$ and $4x - y = 1$.
3. What happens if two equations in a system are parallel?
4. How do you solve a system using the substitution method?
5. Solve the system: $x - 3y = -9$ and $2x + y = 7$.

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