Let's evaluate whether the given ordered pairs satisfy the system of equations:

The system of equations is:

1. $3x - 5y = 1$
2. $2x + 6y = 10$

For the ordered pair $(2, 1)$:

1. Substitute $x = 2$ and $y = 1$ into the first equation: $3(2) - 5(1) = 6 - 5 = 1 \quad \text{(True)}$
2. Substitute $x = 2$ and $y = 1$ into the second equation: $2(2) + 6(1) = 4 + 6 = 10 \quad \text{(True)}$

Since both equations are satisfied, $(2, 1)$ is a solution to the system.

For the ordered pair $(-3, -2)$:

1. Substitute $x = -3$ and $y = -2$ into the first equation: $3(-3) - 5(-2) = -9 + 10 = 1 \quad \text{(True)}$
2. Substitute $x = -3$ and $y = -2$ into the second equation: $2(-3) + 6(-2) = -6 - 12 = -18 \quad \text{(False)}$

Since the second equation is not satisfied, $(-3, -2)$ is not a solution to the system.

Final Answer:

• $(2, 1)$: Yes
• $(-3, -2)$: No

Let me know if you want more details about this process!

Related Questions:

1. What is the graphical interpretation of solving a system of equations like this one?
2. How can substitution or elimination methods be used to solve systems of equations?
3. What happens if both equations represent the same line? How does this affect the solutions?
4. How can matrices and determinants be used to solve systems of equations?
5. What are the implications if a system of equations has no solution or infinitely many solutions?

Tip:

Always check each equation in the system when verifying if a point is a solution. Both equations must be satisfied for it to be valid.