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We are given the following system of linear equations:

\[
6x - y = 16
\]
\[
5x + 3y = 21
\]

To solve this system, we can use either substitution or elimination. Let's use the **substitution method** by solving the first equation for \( y \) and then substituting into the second equation.

1. Solve \( 6x - y = 16 \) for \( y \):
   \[
   y = 6x - 16
   \]

2. Substitute \( y = 6x - 16 \) into the second equation \( 5x + 3y = 21 \):
   \[
   5x + 3(6x - 16) = 21
   \]
   Simplifying:
   \[
   5x + 18x - 48 = 21
   \]
   Combine like terms:
   \[
   23x - 48 = 21
   \]
   Add 48 to both sides:
   \[
   23x = 69
   \]
   Solve for \( x \):
   \[
   x = \frac{69}{23} = 3
   \]

3. Now substitute \( x = 3 \) back into \( y = 6x - 16 \) to find \( y \):
   \[
   y = 6(3) - 16 = 18 - 16 = 2
   \]

Thus, the solution to the system is:
\[
x = 3, \quad y = 2
\]

Would you like more details or explanations on any part of the solution?

Here are five related questions to further explore:
1. What is the elimination method for solving systems of equations?
2. How can we verify the solution to a system of linear equations?
3. Can systems of equations have no solution, and how do we recognize this?
4. What does it mean when a system of equations has infinitely many solutions?
5. How would the graph of these two equations look?

**Tip:** Always substitute your solution back into both original equations to check your answer!

Find the solution(s), if any, of the system:
6x - y = 16
5x + 3y = 21
If there are no solutions, enter 'NS'. If there are infinitely many solutions, enter 'IS'.

Learn how to solve the system of linear equations 6x - y = 16 and 5x + 3y = 21 using the substitution method. This step-by-step solution walks through solving for y and substituting it into the second equation. Perfect for students in Grades 8-10 studying algebra.

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