Question

Which of the following systems of equations has the solution (-2, 1)? This question requires you to show your work. (1 point) Responses 2x−y=−5 and x+2y=10 2 x minus y is equal to negative 5 and x plus 2 y is equal to 10 y=−32x−2 and 2x−y=−5 y is equal to negative 3 over 2 x minus 2 and 2 x minus y is equal to negative 5 y=−2x−2 and 2x−2y=5 y is equal to negative 2 x minus 2 and 2 x minus 2 y is equal to 5 x−4y=8 and 3x+10=−2y

Answer 1

To determine which system of equations has the solution (-2, 1), we will substitute \(x = -2\) and \(y = 1\) into each of the provided pairs of equations.

1. **First System:**
\(2x - y = -5\) and \(x + 2y = 10\)

- For the first equation:
\[
2(-2) - 1 = -4 - 1 = -5 \quad \text{(True)}
\]
- For the second equation:
\[
-2 + 2(1) = -2 + 2 = 0 \quad \text{(Not true, needs to equal 10)}
\]

**Not the solution.**

2. **Second System:**
\(y = -\frac{3}{2}x - 2\) and \(2x - y = -5\)

- For the first equation:
\[
1 = -\frac{3}{2}(-2) - 2 \Rightarrow 1 = 3 - 2 = 1 \quad \text{(True)}
\]
- For the second equation:
\[
2(-2) - 1 = -4 - 1 = -5 \quad \text{(True)}
\]

**This is a valid solution.**

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

- For the first equation:
\[
1 = -2(-2) - 2 \Rightarrow 1 = 4 - 2 = 2 \quad \text{(Not true)}
\]

**Not the solution.**

4. **Fourth System:**
\(x - 4y = 8\) and \(3x + 10 = -2y\)

- For the first equation:
\[
-2 - 4(1) = -2 - 4 = -6 \quad \text{(Not true, needs to equal 8)}
\]

**Not the solution.**

**Conclusion**: The second system, \(y = -\frac{3}{2}x - 2\) and \(2x - y = -5\), is the only one that has the solution (-2, 1).