Question

The point (0,-2) is a solution to which system? -5x + y = -2 and -3x + 6y = -12; -5x = y - 3 and 3x - 8y = 24; -4x + y = 6 and -5x - y = 21; x + y = 2 and -x + 2y = 16

Answer 1

To determine which system has (0, -2) as a solution, we will substitute \(x = 0\) and \(y = -2\) into each system of equations.

1. **System 1:**
\(-5x + y = -2\) and \(-3x + 6y = -12\)

- First equation:
\[
-5(0) + (-2) = -2 \quad (0 - 2 = -2) \quad \text{True}
\]

- Second equation:
\[
-3(0) + 6(-2) = -12 \quad (0 - 12 = -12) \quad \text{True}
\]

Both equations are satisfied by (0, -2).

2. **System 2:**
\(-5x = y - 3\) and \(3x - 8y = 24\)

- First equation:
\[
-5(0) = -2 - 3 \quad (0 = -5) \quad \text{False}
\]

- Second equation is irrelevant since the first equation is already false.

3. **System 3:**
\(-4x + y = 6\) and \(-5x - y = 21\)

- First equation:
\[
-4(0) + (-2) = 6 \quad (0 - 2 = 6) \quad \text{False}
\]

- Second equation is irrelevant since the first equation is already false.

4. **System 4:**
\(x + y = 2\) and \(-x + 2y = 16\)

- First equation:
\[
0 + (-2) = 2 \quad (-2 = 2) \quad \text{False}
\]

- Second equation is irrelevant since the first equation is already false.

After evaluating all four systems, we find that **System 1** is the only one that has (0, -2) as a solution.

**Answer:** System 1: -5x + y = -2 and -3x + 6y = -12.