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.

Learn how to solve the system of linear equations 4x + 3y = 8 and 4x + 6y = 20 using the elimination method. This problem focuses on algebraic concepts, including linear equations and elimination, and is suitable for students in Grades 9-12.

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