We are given the following system of equations:

1. $y = 4x + 1$
2. $3x + 2y = 13$

Step 1: Substitute $y$ from the first equation into the second equation.

Since $y = 4x + 1$, substitute this into the second equation $3x + 2y = 13$:

$3x + 2(4x + 1) = 13$

Step 2: Simplify the equation.

Distribute the $2$ on the left-hand side:

$3x + 8x + 2 = 13$

Combine like terms:

$11x + 2 = 13$

Step 3: Solve for $x$.

Subtract 2 from both sides:

$11x = 11$

Now, divide by 11:

$x = 1$

Step 4: Substitute $x = 1$ back into the first equation to find $y$.

Using $y = 4x + 1$:

$y = 4(1) + 1 = 5$

Final Answer:

The solution to the system of equations is $x = 1$ and $y = 5$.

Would you like more details, or have any questions?

Here are 5 related questions for further exploration:

1. What would happen if the coefficient of $y$ in the second equation changed?
2. Can you solve the system using the substitution method in a different way?
3. How would you approach solving the system graphically?
4. What are some examples of systems of equations that have no solutions?
5. Can you solve a system of equations using matrix methods?

Tip: When solving systems of equations, substitution is often most useful when one equation is already solved for a variable, while elimination is better when variables are aligned for easy elimination.