Question

The solution to a system of linear equations is (negative 3, negative 3). Which system of linear equations has this point as its solution?
x minus 5 y = negative 12 and 3 x + 2 y = negative 15
x minus 5 y = negative 12 and 3 x + 2 y = 15
x minus 5 y = 12 and 3 x + 2 y = negative 15
x minus 5 y = 12 and 3 x + 2 y = 15

Answer 1

To determine which system of linear equations has the solution \((-3, -3)\), we will substitute \(x = -3\) and \(y = -3\) into each of the given options and check which one holds true.

1. **Option 1:** \(x - 5y = -12\) and \(3x + 2y = -15\)

Substitute \(x = -3\) and \(y = -3\):
- First equation:
\[
-3 - 5(-3) = -3 + 15 = 12 \quad \text{(not true)}
\]
- Second equation:
\[
3(-3) + 2(-3) = -9 - 6 = -15 \quad \text{(true)}
\]

Since the first equation is not satisfied, this system does not work.

2. **Option 2:** \(x - 5y = -12\) and \(3x + 2y = 15\)

Substitute \(x = -3\) and \(y = -3\):
- First equation:
\[
-3 - 5(-3) = -3 + 15 = 12 \quad \text{(not true)}
\]
- Second equation:
\[
3(-3) + 2(-3) = -9 - 6 = -15 \quad \text{(not true)}
\]

This system does not work either.

3. **Option 3:** \(x - 5y = 12\) and \(3x + 2y = -15\)

Substitute \(x = -3\) and \(y = -3\):
- First equation:
\[
-3 - 5(-3) = -3 + 15 = 12 \quad \text{(true)}
\]
- Second equation:
\[
3(-3) + 2(-3) = -9 - 6 = -15 \quad \text{(true)}
\]

Both equations are satisfied, so this system works.

4. **Option 4:** \(x - 5y = 12\) and \(3x + 2y = 15\)

Substitute \(x = -3\) and \(y = -3\):
- First equation:
\[
-3 - 5(-3) = -3 + 15 = 12 \quad \text{(true)}
\]
- Second equation:
\[
3(-3) + 2(-3) = -9 - 6 = -15 \quad \text{(not true)}
\]

This system does not work.

Thus, the only system that has the solution \((-3, -3)\) is **Option 3:** \(x - 5y = 12\) and \(3x + 2y = -15\).