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The system of equations provided is:

\[
\begin{cases}
x - 3y = 5 \\
2x + y = 3
\end{cases}
\]

Let's solve this system using the substitution or elimination method. I will proceed with substitution.

1. From the second equation \( 2x + y = 3 \), solve for \( y \):
\[
y = 3 - 2x
\]

2. Substitute \( y = 3 - 2x \) into the first equation \( x - 3y = 5 \):
\[
x - 3(3 - 2x) = 5
\]
\[
x - 9 + 6x = 5
\]
\[
7x - 9 = 5
\]
\[
7x = 14
\]
\[
x = 2
\]

3. Now substitute \( x = 2 \) back into the equation \( y = 3 - 2x \):
\[
y = 3 - 2(2) = 3 - 4 = -1
\]

Thus, the solution is:
\[
x = 2, \quad y = -1
\]

Would you like more details or have any questions?

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Here are 5 related questions:

1. How can you verify the solution of a system of equations?
2. Can this system be solved using the elimination method? How?
3. What is the graphical interpretation of solving a system of linear equations?
4. How does the substitution method differ from the elimination method in solving systems?
5. Can this system be solved using matrix methods such as Gaussian elimination?

**Tip:** Always check your solution by substituting the values of \( x \) and \( y \) back into the original equations.

Solve the system of linear equations: x - 3y = 5 and 2x + y = 3.

Learn how to solve the system of linear equations x - 3y = 5 and 2x + y = 3 using the substitution method. This step-by-step solution is perfect for students in Grades 9-10 and covers essential algebraic concepts for solving systems of equations.

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