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Direct Variation
What is it and how do I know when I see it?
Unit 8 – Vocabulary
 The terms DIRECT VARIATION and
DIRECT PROPORTION will be used
interchangeably. They mean the
same thing.
 The terms CONSTANT OF
PROPORTIONALITY and CONSTANT
OF VARIATION will be used
interchangeably. They mean the
same thing.
What is a Direct Variation or a Direct
Proportion?
When one value increases, the
other value also increases OR
When one value decreases, the
other value also decreases
When one quantity always
changes by the same factor (the
constant of proportionality), the
two quantities are directly
proportional.
Real World Examples of Direct
Variation Situations…
 The more time I drive (at a constant rate),
the more miles I go.
 If I increase a recipe for more people, the
more of an ingredient I need.
 The more hours I work, the more money I
make.
 The more CD’s I purchase, the more
money it costs.
 The less cheese I buy at the deli, the less
money I pay.
 The less water you drink, the less trips to
the bathroom you have to make.
Now you come up with a few
examples of your own…
Raise your hand if you have an
example of when both values
would increase
Raise your hand if you have an
example of when both values
would decrease
Choose one of each to write in
your notes
How do we know if we have a direct
variation?
We can look at three different
aspects of the situation:
1. TABLES
2. EQUATIONS
3. GRAPHS
Direct Variation Equations:
• Y varies directly as x means that y = kx
where k is the constant of variation.
• Another way of writing this is k =
• X is the independent variable
• Y is the dependent variable
• K is the constant of proportionality
y
x
Independent VS. Dependent
The x-axis is the independent
variable; this means it does NOT
depend on the y value
The y-axis is the dependent
variable; this means it DOES
depend on the x variable for its
value.
Direct Variation & Tables of Values
You can make a table of values for “x”
and “y” and see how the values
behave. You could have a direct
variation if…
 As “x” increases in value, “y” also
increases in value
OR
 As “x” decreases in value, “y” also
decreases in value
Examples of Direct Variation:
X Y
6 12
7 14
8 16
Note: As “x” increases,
6 , 7 , 8
“y” also increases.
12, 14, 16
What is the constant of variation of the table above?
Start with the direct variation equation: y = kx
Pick one pair of x and y values and substitute into the equation
12 = k · 6 (this is a one-step equation, so solve for k)
12/6 = k → k = 2
Now you can write the equation for this direct variation: y = 2x
Examples of Direct Variation
Equations (y = kx or y/x = k)…
y = 4x  k = 4
y = x  k = 1
y = 2x  k = 2
y = 2.5x  k = 2.5
y = ⅝ x  k = ⅝
y = 0.75  k = 0.75
x
X Y
30 10
15 5
9 3
Note: X decreases,
30, 15, 9
And Y decreases.
10, 5, 3
What is the constant of variation of the table above?
Start with the direct variation equation: y = kx
Pick one pair of x and y values and substitute into the equation
10 = k · 30 (this is a one-step equation, so solve for k)
10/30 = k → (simplify 10/30) → k = ⅓
Now you can write the equation for this direct variation: y = ⅓ · x
Examples of Direct Variation:
Is this a direct variation? If yes, give the
constant of variation (k) and the equation.
X Y
4 6
8 12
12 18
18 27
Yes!
y = kx
•Pick an x & y pair and
substitute into the direct
variation equation to solve for k.
•Remember the constant must
hold true for every (x,y) pair
6 = k · 4
k = 6/4 = 3/2 = 1 ½
Therefore the equation for this
table is: y = 1 ½ · x
X Y
10 25
6 15
4 10
2 5
Yes!
y = kx
* Pick an x & y pair and
substitute into the direct
variation equation & find k.
25 = k · 10
25/10 = k
5/2 or 2 ½ = k
* Remember the constant must
hold true for every x,y pair.
Therefore the equation for this
table is:
y = 2 ½ · x
Is this a direct variation? If yes, give the
constant of variation (k) and the equation.
X Y
15 5
3 26
1 75
2 150
No!
If you look at the values in
the table, you should notice
as “x” decreases, “y”
increases, so you know you
CANNOT have a direct
variation!
Also, there is no constant of
proportionality. There is not
one number you multiply by
x to get y for each pair in the
table.
Is this a direct variation? If yes, give the
constant of variation (k) and the equation.
Which is the equation that describes
the following table of values?
X Y
10 5
2 1
12 6
20 10
1. y = -2x
2. y = 2x
3. y = ½ x
4. xy = 200
Answer
Now
Using Direct Variation to find unknowns (y = kx)
Given y varies directly with x, and y = 28 when x=7,
Find x when y = 52. HOW???
2 step process
X Y
7 28
? 52
1. Find the constant of variation
y = kx → 28 = k · 7
(divide both sides by 7)
k=4
2. Use y = kx. Find the unknown (x).
52= 4x or 52/4 = x
x= 13
Therefore:
X =13 when Y=52
Using Direct Variation to solve word problems
Problem:
A car uses 8 gallons of
gasoline to travel 290
miles. How much
gasoline will the car use
to travel 400 miles?
Step One: Find points in table
X (gas) Y (miles)
8 290
? 400
Step Two: Find the constant
of variation and equation:
y = kx → 290 = k · 8
290/8 = k
y = 36.25 x
Step Three: Use the equation
to find the unknown.
400 =36.25x
400 =36.25x
36.25 36.25
or x = 11.03
Given that y varies directly with x, and y = 3 when x=9,
Find y when x = 40.5. HOW???
2 step process
X Y
9 3
40.5 ?
1. Find the constant of variation.
Y = kx → 3 = k · 9
(divide both sides by 9)
k = 3/9 = 1/3
2. Use y = kx. Find the unknown (x).
y= (1/3)40.5
y= 13.5
Therefore:
X =40.5 when
Y=13.5
Using Direct Variation to find unknowns (y = kx)
Using Direct Variation to solve word problems
Problem:
Julio wages vary
directly as the number
of hours that he works.
If his wages for 5 hours
are $29.75, how much
will they be for 30 hours
Step One: Find points in table.
X(hours) Y(wages)
5 29.75
30 ?
Step Two: Find the constant of
variation.
y = kx → 29.75 = k · 5
k = 5.95
Step Three: Use the equation
to find the unknown.
y = kx
y = 5.95 ·30
y = 178.50
Direct Variation and Its Graph
Characteristics of Direct Proportion
Graph…
 The graph will always go through the
ORIGIN (point 0,0) on the coordinate
plane…this means when x=0, y=0 on the
graph
 The graph will always be in Quadrant I (all
positive numbers) and Quadrant III (all
negative numbers – this is for next year)
 The graph will always be a straight line
 As the “x” value increases, the “y” value
always increases
These things will always occur in DIRECT
VARIATIONS…if it doesn’t then it is NOT a
DIRECT PROPORTION!
Tell if the following graph is a Direct
Variation or not.
No
No
No No
No Yes
Yes No
Tell if the following graph is a Direct
Variation or not.
Direct Variation/Direct
Proportion Reminder…
 When you have word problems, you
can also set up proportions to solve
for the given situation if you prefer.
 If you’re going to set up a
proportion, it’s best to put the y-
value (the dependent value) over the
x-value (the independent value)
because then it’s already set up to
solve for “k” the constant of
proportionality.

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Direct variation-ppt

  • 1. Direct Variation What is it and how do I know when I see it?
  • 2. Unit 8 – Vocabulary  The terms DIRECT VARIATION and DIRECT PROPORTION will be used interchangeably. They mean the same thing.  The terms CONSTANT OF PROPORTIONALITY and CONSTANT OF VARIATION will be used interchangeably. They mean the same thing.
  • 3. What is a Direct Variation or a Direct Proportion? When one value increases, the other value also increases OR When one value decreases, the other value also decreases When one quantity always changes by the same factor (the constant of proportionality), the two quantities are directly proportional.
  • 4. Real World Examples of Direct Variation Situations…  The more time I drive (at a constant rate), the more miles I go.  If I increase a recipe for more people, the more of an ingredient I need.  The more hours I work, the more money I make.  The more CD’s I purchase, the more money it costs.  The less cheese I buy at the deli, the less money I pay.  The less water you drink, the less trips to the bathroom you have to make.
  • 5. Now you come up with a few examples of your own… Raise your hand if you have an example of when both values would increase Raise your hand if you have an example of when both values would decrease Choose one of each to write in your notes
  • 6. How do we know if we have a direct variation? We can look at three different aspects of the situation: 1. TABLES 2. EQUATIONS 3. GRAPHS
  • 7. Direct Variation Equations: • Y varies directly as x means that y = kx where k is the constant of variation. • Another way of writing this is k = • X is the independent variable • Y is the dependent variable • K is the constant of proportionality y x
  • 8. Independent VS. Dependent The x-axis is the independent variable; this means it does NOT depend on the y value The y-axis is the dependent variable; this means it DOES depend on the x variable for its value.
  • 9. Direct Variation & Tables of Values You can make a table of values for “x” and “y” and see how the values behave. You could have a direct variation if…  As “x” increases in value, “y” also increases in value OR  As “x” decreases in value, “y” also decreases in value
  • 10. Examples of Direct Variation: X Y 6 12 7 14 8 16 Note: As “x” increases, 6 , 7 , 8 “y” also increases. 12, 14, 16 What is the constant of variation of the table above? Start with the direct variation equation: y = kx Pick one pair of x and y values and substitute into the equation 12 = k · 6 (this is a one-step equation, so solve for k) 12/6 = k → k = 2 Now you can write the equation for this direct variation: y = 2x
  • 11. Examples of Direct Variation Equations (y = kx or y/x = k)… y = 4x  k = 4 y = x  k = 1 y = 2x  k = 2 y = 2.5x  k = 2.5 y = ⅝ x  k = ⅝ y = 0.75  k = 0.75 x
  • 12. X Y 30 10 15 5 9 3 Note: X decreases, 30, 15, 9 And Y decreases. 10, 5, 3 What is the constant of variation of the table above? Start with the direct variation equation: y = kx Pick one pair of x and y values and substitute into the equation 10 = k · 30 (this is a one-step equation, so solve for k) 10/30 = k → (simplify 10/30) → k = ⅓ Now you can write the equation for this direct variation: y = ⅓ · x Examples of Direct Variation:
  • 13. Is this a direct variation? If yes, give the constant of variation (k) and the equation. X Y 4 6 8 12 12 18 18 27 Yes! y = kx •Pick an x & y pair and substitute into the direct variation equation to solve for k. •Remember the constant must hold true for every (x,y) pair 6 = k · 4 k = 6/4 = 3/2 = 1 ½ Therefore the equation for this table is: y = 1 ½ · x
  • 14. X Y 10 25 6 15 4 10 2 5 Yes! y = kx * Pick an x & y pair and substitute into the direct variation equation & find k. 25 = k · 10 25/10 = k 5/2 or 2 ½ = k * Remember the constant must hold true for every x,y pair. Therefore the equation for this table is: y = 2 ½ · x Is this a direct variation? If yes, give the constant of variation (k) and the equation.
  • 15. X Y 15 5 3 26 1 75 2 150 No! If you look at the values in the table, you should notice as “x” decreases, “y” increases, so you know you CANNOT have a direct variation! Also, there is no constant of proportionality. There is not one number you multiply by x to get y for each pair in the table. Is this a direct variation? If yes, give the constant of variation (k) and the equation.
  • 16. Which is the equation that describes the following table of values? X Y 10 5 2 1 12 6 20 10 1. y = -2x 2. y = 2x 3. y = ½ x 4. xy = 200 Answer Now
  • 17. Using Direct Variation to find unknowns (y = kx) Given y varies directly with x, and y = 28 when x=7, Find x when y = 52. HOW??? 2 step process X Y 7 28 ? 52 1. Find the constant of variation y = kx → 28 = k · 7 (divide both sides by 7) k=4 2. Use y = kx. Find the unknown (x). 52= 4x or 52/4 = x x= 13 Therefore: X =13 when Y=52
  • 18. Using Direct Variation to solve word problems Problem: A car uses 8 gallons of gasoline to travel 290 miles. How much gasoline will the car use to travel 400 miles? Step One: Find points in table X (gas) Y (miles) 8 290 ? 400 Step Two: Find the constant of variation and equation: y = kx → 290 = k · 8 290/8 = k y = 36.25 x Step Three: Use the equation to find the unknown. 400 =36.25x 400 =36.25x 36.25 36.25 or x = 11.03
  • 19. Given that y varies directly with x, and y = 3 when x=9, Find y when x = 40.5. HOW??? 2 step process X Y 9 3 40.5 ? 1. Find the constant of variation. Y = kx → 3 = k · 9 (divide both sides by 9) k = 3/9 = 1/3 2. Use y = kx. Find the unknown (x). y= (1/3)40.5 y= 13.5 Therefore: X =40.5 when Y=13.5 Using Direct Variation to find unknowns (y = kx)
  • 20. Using Direct Variation to solve word problems Problem: Julio wages vary directly as the number of hours that he works. If his wages for 5 hours are $29.75, how much will they be for 30 hours Step One: Find points in table. X(hours) Y(wages) 5 29.75 30 ? Step Two: Find the constant of variation. y = kx → 29.75 = k · 5 k = 5.95 Step Three: Use the equation to find the unknown. y = kx y = 5.95 ·30 y = 178.50
  • 21. Direct Variation and Its Graph
  • 22. Characteristics of Direct Proportion Graph…  The graph will always go through the ORIGIN (point 0,0) on the coordinate plane…this means when x=0, y=0 on the graph  The graph will always be in Quadrant I (all positive numbers) and Quadrant III (all negative numbers – this is for next year)  The graph will always be a straight line  As the “x” value increases, the “y” value always increases These things will always occur in DIRECT VARIATIONS…if it doesn’t then it is NOT a DIRECT PROPORTION!
  • 23. Tell if the following graph is a Direct Variation or not. No No No No
  • 24. No Yes Yes No Tell if the following graph is a Direct Variation or not.
  • 25. Direct Variation/Direct Proportion Reminder…  When you have word problems, you can also set up proportions to solve for the given situation if you prefer.  If you’re going to set up a proportion, it’s best to put the y- value (the dependent value) over the x-value (the independent value) because then it’s already set up to solve for “k” the constant of proportionality.