To determine which points are solutions to the system of equations \(y = \frac{1}{2}x - 2\) and \(y = \frac{3}{2}x - 4\), we can substitute each point into both equations and see if they satisfy both.
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Point (2, -1):
- For \(y = \frac{1}{2}(2) - 2\): \[ y = 1 - 2 = -1 \quad (\text{True}) \]
- For \(y = \frac{3}{2}(2) - 4\): \[ y = 3 - 4 = -1 \quad (\text{True}) \]
- Point (2, -1) is a solution.
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Point (-2, -3):
- For \(y = \frac{1}{2}(-2) - 2\): \[ y = -1 - 2 = -3 \quad (\text{True}) \]
- For \(y = \frac{3}{2}(-2) - 4\): \[ y = -3 - 4 = -7 \quad (\text{False}) \]
- Point (-2, -3) is not a solution.
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Point (1, -1.5):
- For \(y = \frac{1}{2}(1) - 2\): \[ y = 0.5 - 2 = -1.5 \quad (\text{True}) \]
- For \(y = \frac{3}{2}(1) - 4\): \[ y = 1.5 - 4 = -2.5 \quad (\text{False}) \]
- Point (1, -1.5) is not a solution.
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Point (-1, -2.5):
- For \(y = \frac{1}{2}(-1) - 2\): \[ y = -0.5 - 2 = -2.5 \quad (\text{True}) \]
- For \(y = \frac{3}{2}(-1) - 4\): \[ y = -1.5 - 4 = -5.5 \quad (\text{False}) \]
- Point (-1, -2.5) is not a solution.
Conclusion: The only point that is a solution to the system of equations is (2, -1).