To solve the system of linear equations:

$-x + 2y = 13 \quad (1)$ $5x + 3y = 13 \quad (2)$

We can use either substitution or elimination. Let's use substitution in this case:

Step 1: Solve Equation (1) for $x$.

From equation (1): $-x = 13 - 2y \quad \Rightarrow \quad x = 2y - 13$

Step 2: Substitute this expression for $x$ in Equation (2).

Substitute $x = 2y - 13$ into equation (2): $5(2y - 13) + 3y = 13$ Expand the equation: $10y - 65 + 3y = 13$ Simplify: $13y - 65 = 13$ Add 65 to both sides: $13y = 78$ Solve for $y$: $y = \frac{78}{13} = 6$

Step 3: Substitute $y = 6$ back into the expression for $x$.

Now, substitute $y = 6$ into the equation $x = 2y - 13$: $x = 2(6) - 13 = 12 - 13 = -1$

Final Solution:

The solution to the system is: $x = -1, \quad y = 6$

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

Here are 5 related questions for further practice:

1. Solve the system: $3x + 2y = 7$ and $2x - y = 4$.
2. Solve the system: $2x - y = 5$ and $-3x + 4y = -6$.
3. What is the graphical interpretation of the solution $x = -1$, $y = 6$?
4. Use the elimination method to solve: $4x + 3y = 18$ and $6x - 5y = 12$.
5. Solve a 3-variable system: $x + y + z = 10$, $2x - y + z = 5$, and $3x + 2y - z = 15$.

Tip: Always double-check your solution by substituting the values of $x$ and $y$ back into both original equations.