line l passes through (4,31) and line p is the graph of y+4=5(x4).
Algebra

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line l passes through (4,31) and line p is the graph of y+4=5(x4). If l is parallel to p what is the slopeintercept form equation of l?
y  y1 = m(x  x1) PointSlope form
y + 4 = 5(x  4) m is the slope
By comparison, m = 5
Line L and Line P are parallel if they have the same slope. And since Line P has a slope of 5, that means that Line L also has a slope of 5. Therefore the slope of Line L is m = 5.
Line L also has the given point of (4, 31). This means we can let (x1, y1) = (4, 31), or rather x1 = 4 and y1 = 31. With the point and slope of Line L now known, we can plug back into the PointSlope form. Keep in mind that we are instructed to put the equation into SlopeIntercept form y = mx + b, meaning that after we substitute, we would rearrange the equation to put in the form y = mx + b (or rather as an equation of y in terms of x)
y  y1 = m(x  x1)
y  (31) = 5(x  4) Substitute the point and slope of Line L
y + 31 = 5(x  4) Simplify if necessary, then convert to y = mx + b
y + 31 = 5x + 20 Distribute the parenthesis
y + 31  31 = 5x + 20  31 Get the y by itself
y = 5x  11 Once y is isolated, it is in slopeintercept form y = mx + b
SOLUTION:
SlopeIntercept equation of Line L is y = 5x  11
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