This document defines and provides examples of direct variation, also called direct proportion. It states that direct variation occurs when one quantity changes by the same factor as another quantity. The constant of proportionality or variation describes this fixed factor of change between the quantities. Examples are provided of direct variation situations in tables, equations, and graphs. The key characteristics of a direct variation graph are that it passes through the origin and yields increasing y-values as x increases. The document provides instructions for identifying and solving direct variation problems using tables, equations, and setting up proportions.

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.