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INSTRUCTORS Carleen Eaton Grant Fraser Eric Smith
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For more information, please see full course syllabus of Algebra 1
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Direct Variation

  • A direct variation is given by an equation of the form y = kx, where k is a nonzero constant, called the constant of variation.
  • The graph of a direct variation is a straight line with slope k and passing through the origin.
  • You can solve problems involving direct variation when you are given values satisfied by the direct variation and asked to find a missing value. First, use the given values to find k. Then use k to find the unknown value.

Direct Variation

If y varies directly with x and if y = 15 when x = 3, find x when y = 25.
  • y = kx
  • 15 = 3k
  • k = 5
  • When y = 25
    25 = 5x
x = 5
If y varies directly with x and if y = 24 when x = 8, find x when y = 36.
  • y = kx
  • 24 = 8k
  • k = 3
  • When y = 36
    36 = 3x
x = 12
If y varies directly with x and if y = 16 when x = 2, find x when y = 22.
  • y = kx
  • 16 = 2k
  • k = 8
  • When y = 22
    22 = 8x
  • [22/8] = x
[11/4] = x
Graph the direct variation given by the equation y = − [1/2]x
  • Find the k value for the slope
  • k = − [1/2]
Graph from the origin
includegraphicsALG-IMG-4-2-4.jpg
Graph the direct variation given by the equation y = 4x
  • Find the k value for the slope
  • k = 4
Graph from the origin
Determine the direct variation equation by the graph given
  • Determine the k value
    k = [(6 − 0)/(5 − 0)] = [6/5]
  • Apply to the direct variation formula
y = kx = [6/5]x
Determine the direct variation equation by the graph given
  • Determine the k value
    k = [(0 − 9)/(0 − ( − 8))] = − [9/8]
  • Apply to the direct variation formula
y = kx = − [9/8]x
Graph the following : suppose y varies directly with x, and y = 28 when x = 7
  • Use the direct variation equation
    y = kx
  • 28 = 7k
  • k = 4
Graph equation
Suppose that a car travels 180 miles in 4 hours. Determine its direct variation equation and graph it as well.
  • Use the direct variation formula in the context of distance and time
    y = kxdistance = time(k)
  • 180 = 4k
  • k = [180/4] = 45
  • y = 45x
Graph the equation
Suppose the same car from the previous problem. How long will it take to make 500 miles?
  • Utilize direct variation equation
  • 500 = 45x
x = [500/45]  ∼ 11.11 hours

*These practice questions are only helpful when you work on them offline on a piece of paper and then use the solution steps function to check your answer.

Answer

Direct Variation

Lecture Slides are screen-captured images of important points in the lecture. Students can download and print out these lecture slide images to do practice problems as well as take notes while watching the lecture.

  • Intro 0:00
  • Definitions 0:10
    • Constant of Variation k
    • Example: Gas and Miles Driven
  • Graph 1:50
    • k is Slope
    • Examples
  • Applications 2:47
    • Write, Graph, Solve
  • Example 1: Constant of Variation 3:11
  • Example 2: Graph Direct Variation 4:59
  • Example 3: Direct Variation 6:50
  • Example 4: Distance Car Travels 9:18