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The image contains a system of three linear equations with three variables \( r \), \( s \), and \( t \):

\[
\begin{aligned}
1. \quad 6r - s + 3t &= -9 \\
2. \quad 5r + 5s - 5t &= 20 \\
3. \quad 3r - s + 4t &= -5
\end{aligned}
\]

I will solve this system of equations step by step using elimination or substitution methods. Let's begin by simplifying and solving for \( r \), \( s \), and \( t \).The solution to the system of equations is:

\[
r = -1, \quad s = 6, \quad t = 1
\]

Would you like a detailed explanation of the steps used to solve this, or do you have any further questions?

Here are 5 related questions for further practice:
1. How can substitution be used to solve a system of three equations?
2. What is the elimination method, and when is it preferred?
3. How do you interpret the solution of a system of equations geometrically?
4. What happens when a system of equations has no solution or infinitely many solutions?
5. How can systems of equations be solved using matrices or determinants?

**Tip**: When solving systems of equations, it is often helpful to align the variables and constants before applying any method of solving.

Solve the system of linear equations:
1) 6r - s + 3t = -9
2) 5r + 5s - 5t = 20
3) 3r - s + 4t = -5

Learn how to solve a system of three linear equations with three variables (r, s, t) using elimination and substitution methods. This example covers solving the system step-by-step and is suitable for students in Grades 9-12. Key concepts include algebra and linear systems.

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