Let's solve the system of linear equations:

1. \( x + 4y = 6 \)
2. \( x - y = 1 \)

### Step 1: Solve one of the equations for one variable
From the second equation, \( x - y = 1 \), solve for \( x \):
\[
x = y + 1
\]

### Step 2: Substitute into the first equation
Now, substitute \( x = y + 1 \) into the first equation \( x + 4y = 6 \):
\[
(y + 1) + 4y = 6
\]
Simplify the equation:
\[
y + 1 + 4y = 6
\]
\[
5y + 1 = 6
\]
\[
5y = 5
\]
\[
y = 1
\]

### Step 3: Solve for \( x \)
Now substitute \( y = 1 \) back into the equation \( x = y + 1 \):
\[
x = 1 + 1 = 2
\]

### Final Solution:
\[
x = 2, \quad y = 1
\]

Would you like further details or have any questions?

Here are five related questions:
1. How do you graph this system of equations?
2. What is the significance of the solution in terms of the lines?
3. Can a system of linear equations have no solution or infinitely many solutions?
4. How do we solve larger systems of equations?
5. What happens if both equations are parallel?

**Tip:** Always check your solution by substituting the values back into the original equations.

Learn how to solve the system of linear equations x + 4y = 6 and x - y = 1 using the substitution method. This step-by-step guide covers important algebraic concepts and is suitable for students in Grades 8-10.

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