You are given the following system of equations:

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

Step 1: Substituting $x = y + 13$ into the first equation

We can substitute $x = y + 13$ from the second equation into the first equation to eliminate $x$.

$4(y + 13) + 2y = 13$

Step 2: Expanding and simplifying the equation

Now, expand and simplify the equation:

$4y + 52 + 2y = 13$

Combine like terms:

$6y + 52 = 13$

Step 3: Solving for $y$

Now, solve for $y$:

$6y = 13 - 52$ $6y = -39$ $y = \frac{-39}{6} = -\frac{13}{2}$

Step 4: Solving for $x$

Now that we know $y = -\frac{13}{2}$, substitute it back into the second equation $x = y + 13$:

$x = -\frac{13}{2} + 13$ $x = -\frac{13}{2} + \frac{26}{2} = \frac{13}{2}$

Final Solution:

The solution to the system of equations is: $x = \frac{13}{2}, \quad y = -\frac{13}{2}$

Would you like further details or have any questions about the process?

Here are some related questions:

1. How can substitution be used in systems of linear equations?
2. What happens if there is no solution or infinite solutions to a system of equations?
3. How does elimination compare to substitution for solving systems?
4. Can you explain how to graph these equations and find the intersection?
5. How do changes to the constants affect the solutions in linear systems?

Tip: When substituting one equation into another, always simplify the equation carefully to avoid errors.