To find the equation of a line that is parallel to the line given by the equation \(6x - 4y = -16\) and passes through the point \((-8, 3)\), we first need to find the slope of the given line.
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Convert the equation to slope-intercept form (i.e., \(y = mx + b\)): \[ 6x - 4y = -16 \] Rearranging gives: \[ -4y = -6x - 16 \] Dividing everything by \(-4\): \[ y = \frac{3}{2}x + 4 \] Here, the slope \(m\) is \(\frac{3}{2}\).
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Since parallel lines have the same slope, the slope of our new line will also be \(\frac{3}{2}\).
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Use the point-slope form of a line's equation: The point-slope form is given by: \[ y - y_1 = m(x - x_1) \] where \((x_1, y_1)\) is the point the line passes through, and \(m\) is the slope. Here, \(m = \frac{3}{2}\) and the point is \((-8, 3)\): \[ y - 3 = \frac{3}{2}(x + 8) \]
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Distributing the slope: \[ y - 3 = \frac{3}{2}x + 12 \]
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Add 3 to both sides: \[ y = \frac{3}{2}x + 15 \]
Therefore, the equation of the line parallel to \(6x - 4y = -16\) that passes through the point \((-8, 3)\) is:
\[ \boxed{y = \frac{3}{2}x + 15} \]
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