The given system of equations is:

x + 3y = 1 \tag{1} y = -x + 15 \tag{2}

We can use substitution to solve for $x$. From equation $(2)$, we already have $y$ in terms of $x$, so we can substitute $y = -x + 15$ into equation $(1)$.

Step 1: Substitute equation $(2)$ into equation $(1)$:

$x + 3(-x + 15) = 1$

Step 2: Expand and solve for $x$:

$x - 3x + 45 = 1$ $-2x + 45 = 1$

Step 3: Move the constant to the other side:

$-2x = 1 - 45$ $-2x = -44$

Step 4: Solve for $x$:

$x = \frac{-44}{-2} = 22$

Thus, the solution for $x$ is:

$x = 22$

The correct answer is C. 22.

Would you like further details or clarification on any step?

Here are 5 related questions for further exploration:

1. How would the solution change if the second equation were $y = x + 15$?
2. Can you graph these two equations to visually confirm the solution?
3. What method would you use to solve this system using elimination?
4. How do you interpret the solution in terms of the intersection of two lines?
5. How could this system be solved using matrix methods?

Tip: Always check your solution by substituting the values back into both original equations to ensure consistency.