To determine the system of linear equations represented by the given table of values, we need to identify the equations for \( y_a \) and \( y_b \) based on the \( x \) values provided.
For \( y_a \):
Using the points (0, 3), (1, 5), and (2, 7), we can determine the slope \( m \) and the y-intercept \( b \).
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Calculate the slope \( m \):
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From (0, 3) to (1, 5): \[ m = \frac{5 - 3}{1 - 0} = \frac{2}{1} = 2 \]
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From (1, 5) to (2, 7): \[ m = \frac{7 - 5}{2 - 1} = \frac{2}{1} = 2 \]
Since the slope is consistent, we can use the point-slope form: \[ y - y_1 = m(x - x_1) \]
Using point (0, 3): \[ y - 3 = 2(x - 0) \implies y = 2x + 3 \]
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For \( y_b \):
Using the points (0, -2), (1, 1), and (2, 4), we will also determine the slope \( m \) and y-intercept \( b \).
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Calculate the slope \( m \):
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From (0, -2) to (1, 1): \[ m = \frac{1 - (-2)}{1 - 0} = \frac{3}{1} = 3 \]
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From (1, 1) to (2, 4): \[ m = \frac{4 - 1}{2 - 1} = \frac{3}{1} = 3 \]
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Using the point-slope form again: Using point (0, -2): \[ y - (-2) = 3(x - 0) \implies y = 3x - 2 \]
Thus, the two equations derived from the table are:
- \( y = 2x + 3 \) (for \( y_a \))
- \( y = 3x - 2 \) (for \( y_b \))
Now, let's match this with the response options provided:
- {y=3x+2 y=-2x+3}
- {y=-2x+3 y=-3x-2}
- {y=2x+3 y=3x-2}
- {y=2x-2 y=3x+3}
The correct system of equations that matches our results is:
{y=2x+3 y=3x-2}.