To graph the equations \( y = \frac{1}{2}x - 1 \) (blue line) and \( y = -\frac{1}{2}x + 3 \) (red line), we first need to identify some points for each line.
For the blue line \( y = \frac{1}{2}x - 1 \):
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When \( x = 0 \): \[ y = \frac{1}{2}(0) - 1 = -1 \quad \Rightarrow \quad (0, -1) \]
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When \( x = 2 \): \[ y = \frac{1}{2}(2) - 1 = 0 \quad \Rightarrow \quad (2, 0) \]
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When \( x = 4 \): \[ y = \frac{1}{2}(4) - 1 = 1 \quad \Rightarrow \quad (4, 1) \]
The points on the blue line are \( (0, -1) \), \( (2, 0) \), and \( (4, 1) \).
For the red line \( y = -\frac{1}{2}x + 3 \):
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When \( x = 0 \): \[ y = -\frac{1}{2}(0) + 3 = 3 \quad \Rightarrow \quad (0, 3) \]
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When \( x = 2 \): \[ y = -\frac{1}{2}(2) + 3 = 2 \quad \Rightarrow \quad (2, 2) \]
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When \( x = 4 \): \[ y = -\frac{1}{2}(4) + 3 = 1 \quad \Rightarrow \quad (4, 1) \]
The points on the red line are \( (0, 3) \), \( (2, 2) \), and \( (4, 1) \).
Finding the Intersection Point:
To solve the system of equations, we can set the two equations equal to each other:
\[ \frac{1}{2}x - 1 = -\frac{1}{2}x + 3 \]
To eliminate the fractions, multiply through by 2:
\[ x - 2 = -x + 6 \]
Now combine like terms:
\[ x + x = 6 + 2 \] \[ 2x = 8 \implies x = 4 \]
Now substitute \( x = 4 \) back into either equation to find \( y \):
Using the blue line: \[ y = \frac{1}{2}(4) - 1 = 2 - 1 = 1 \]
Thus, the intersection point is \( (4, 1) \).
Summary:
- Points on Blue Line: \( (0, -1) \), \( (2, 0) \), \( (4, 1) \)
- Points on Red Line: \( (0, 3) \), \( (2, 2) \), \( (4, 1) \)
- Intersection Point: \( (4, 1) \)
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