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Algebra 1 View details

3. Direct Variation and Proportional Relationships

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Chapter 2

3.

Direct Variation and Proportional Relationships

Direct variation occurs where one variable changes directly in relation to another. In simple terms, if one variable increases, the other does too, maintaining a consistent ratio. Direct variation graphs visualize this relationship, showing a straight line passing through the origin. This connection between variables is termed as "proportional." Recognizing this relationship is crucial for tasks like budgeting, scaling designs, or predicting trends. In real-world scenarios, understanding proportional variables can simplify complex problems, making it easier to make informed decisions.

Lesson Settings & Tools

	10 Theory slides
	8 Exercises - Grade E - A
	Each lesson is meant to take 1-2 classroom sessions

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If two quantities are related and one of them varies, then the other quantity will also vary. Depending on the type of relationship they have, the second quantity will vary in a specific way. This lesson will explore the ideas of direct variation and proportional relationships.

Catch-Up and Review

Here are a few recommended readings before getting started with this lesson.

Here are a few practice exercises before getting started with this lesson.

a Consider the table of values.

$x$	$y = 2 x + 1$
$1$	$3$
$2$	A
$3$	B
$4$	$9$
$5$	C

Find the values of A, B, and C.

b Consider the table of values.

$x$	$y = x + 1$
$- 1$	$0$
$0$	$1$
$1$	$2$
$2$	$3$
$3$	$4$

Consider now the following graphs.

Which of the graphs corresponds to the table?

Explore

Moving a Point Along a Line

Consider the following line through the origin. For any point on this line, the ratio of its

y -

coordinate to its

x -

coordinate is always the same.

Does this happen for every line through the origin?

Discussion

Direct Variation

Direct variation, also known as direct proportionality or proportional relationship, occurs when two variables, $x$ and $y,$ have a relationship that forms a linear function passing through the origin where $x \neq = 0$ and $y \neq = 0 .$

$y = k x$

The constant

k

is the constant of variation. It defines the slope of the line. When

k = 0,

the relationship is not in direct variation. In the example below, the constant of variation is

k = 1.5 .

A line y=1.5x with a point on the line that can be moved

The constant of variation may be any real number except

0 .

It is worth noting that the quotient of

y

and

x

is the constant of variation.

y = k x \Leftrightarrow \frac{y}{x} = k

Example

Graphing a Direct Variation

Paulina is selling lemonade to save some money for her summer vacation. For each glass of lemonade she sells, Paulina makes a profit of

$ 0.50 .

She models this situation with a direct variation.

p = 0.5 g

Here,

p

is the profit made when

g

glasses are sold. Also,

0.5

is the constant of variation.

a Consider the following graphs.

Which of the graphs corresponds to the given direct variation?

b What is Paulina's profit if she sells

5

glasses?

Hint

a Draw the graph of the direct variation and compare it with the given graphs.

b Use the graph from Part A.

Solution

a To determine which of the four graphs represents the given direct variation, the graph of

p = 0.5 g

will be drawn by making a table of values.

$g$	$0.5 g$	$p = 0.5 g$
$0$	$0.5 (0)$	$0$
$1$	$0.5 (1)$	$0.5$
$2$	$0.5 (2)$	$1$
$3$	$0.5 (3)$	$1.5$
$4$	$0.5 (4)$	$2$

Now, the obtained points in the table will be plotted and connected with a straight line.

The obtained graph is the same as Graph II. Note that in the context of the situation, negative values of

g

do not make sense, since Paulina cannot sell a negative number of glasses of lemonade.

b To calculate Paulina's profit when selling

5

glasses of lemonade, the graph from Part A will be used.

The profit for

5

glasses is

$ 2.50 .

Note that this can be verified using the equation. To do so,

5

will be substituted for

g .

p = 0.5 g substitute p p = 0.5 (5) = 2.5 ✓

Example

Writing a Direct Variation

Tadeo is given the following math homework.

Help Tadeo do his homework!

Hint

The general form of a direct variation is $y = k x,$ where $k$ is the constant of variation.

Solution

Start by writing the general form of a direct variation.

y = k x

Here,

k

is the constant of variation. To find its value,

x = 16

and

y = 12

will be substituted into the equation.

y = k x

SubstituteII

$x = 16$ , $y = 12$

12 = k (16)

▼

Solve for

k

DivEqn

$LHS / 16 = RHS / 16$

\frac{1 2}{1 6} = k

ReduceFrac

$\frac{a}{b} = \frac{a / 4}{b / 4}$

\frac{3}{4} = k

RearrangeEqn

Rearrange equation

k = \frac{3}{4}

The constant of variation is

\frac{3}{4} .

With this information, the equation can be written.

y = k x substitute y = \frac{3}{4} x

Finally, this equation can be used to find the value of

y

when

x = 24 .

y = \frac{3}{4} x

Substitute

$x = 24$

y = \frac{3}{4} (24)

▼

Evaluate right-hand side

MoveRightFacToNum

$\frac{a}{c} \cdot b = \frac{a \cdot b}{c}$

y = \frac{7 2}{4}

CalcQuot

Calculate quotient

y = 18

When

x = 24,

the value of

y

is

18 .

Example

Writing a Direct Variation Knowing a Point

Tearrik is given the graph of a direct variation and one of its points.

To complete his homework before going dancing, he wants to find the equation of the direct variation shown in the graph. Help him do this!

Hint

The equation of a direct variation is $y = k x,$ where $k$ is the constant of variation.

Solution

Start by recalling the general form of a direct variation.

y = k x

Here,

k

is the constant of variation. To find its value, the point

(4, 5)

will be used. To do this,

x = 4

and

y = 5

will be substituted into the above equation.

y = k x

SubstituteII

$x = 4$ , $y = 5$

5 = k (4)

▼

Solve for

k

DivEqn

$LHS / 4 = RHS / 4$

1.25 = k

RearrangeEqn

Rearrange equation

k = 1.25

The constant of variation is

1.25 .

With this information, the formula can be written.

y = k x substitute y = 1.25 x

Example

Writing a Direct Variation Knowing Its Graph

This time, Tearrik is given the graph of a direct variation, but none of its points are plotted.

Once again, to complete his homework before going dancing, he wants to find the equation of the direct variation shown in the graph. Help him do this!

Hint

Use any point on the line.

Solution

Recall the general form of a direct variation.

y = k x

Here,

k

is the constant of variation. To find its value, any point on the given graph can be used. For simplicity, the point

(- 10, 8)

will be considered.

To find the constant of variation,

x = - 10

and

y = 8

will be substituted into the general formula.

y = k x

SubstituteII

$x = - 10$ , $y = 8$

8 = k (- 10)

▼

Solve for

k

DivEqn

$LHS / (- 10) = RHS / (- 10)$

\frac{8}{- 1 0} = k

MoveNegDenomToFrac

Put minus sign in front of fraction

- \frac{8}{1 0} = k

CalcQuot

Calculate quotient

- 0.8 = k

RearrangeEqn

Rearrange equation

k = - 0.8

The constant of variation is

- 0.8 .

With this information, the equation of the direct variation of the given graph can be written.

y = k x substitute y = - 0.8 x

Pop Quiz

Finding the Constant of Variation

Find the constant of variation of the direct variation whose graph is given. If the answer is not an integer, write it as a decimal rounded to one decimal place.

Example

Modeling With Direct Variation

Tiffaniqua is jogging on Saturday morning. As she jogs, Tiffaniqua keeps a constant speed of

8 km / h .

a Let

d

represent the distance in kilometers and

t

the time in hours. Write a direct variation in terms of

d

and

t

to represent this situation.

b Consider the following graphs.

Which of these is the graph of the direct variation from Part A?

c How many kilometers would Tiffaniqua travel in

45

minutes?

How many hours would it take her to travel

12

kilometers?

d How many kilometers would Tiffaniqua travel in

6

hours and

45

minutes?

How many hours will it take her to travel

44

kilometers?

Hint

a What is the constant of variation?

b If Tiffaniqua's speed is

8 km / h,

then she travels

8

kilometers in one hour.

c Use the graph from Part B.

d Use the formula from Part A.

Solution

a The distance traveled

d

varies directly with the time

t .

Since Tiffaniqua's speed is

8 km / h,

the constant of variation is

8 .

d = k t substitute d = 8 t

b The graph of a direct variation passes through the origin. Therefore, the point

(0, 0)

is on the graph of

d = 8 t .

To find another point on the line, any value can be substituted for

t

in the formula. For simplicity,

t = 1

will be used.

d = 8 t substitute d d = 8 (1) = 8

It has been found that the point

(1, 8)

is also on the line. To draw the graph, these two points will be plotted and the line through them will be drawn.

This graph corresponds to Graph III. Note that in the context of the situation, a negative value of

t

does not make sense, since Tiffaniqua cannot run for a negative amount of hours.

c To find how many kilometers Tiffaniqua would travel in

45

minutes and how many hours it would take her to travel

12

kilometers, the graph from Part B will be used.

It can be seen that, if Tiffaniqua jogged at a constant speed of

8

kilometers per hour, she would travel

6

kilometers in

45

minutes. Similarly, it would take her one and a half hours to travel

12

kilometers.

d Since these two values are too large to be seen in the graph, the formula from Part A will be used.

d = 8 t

Recall that

d

is the distance in kilometers and

t

the time in hours. To find how many kilometers Tiffaniqua would travel in

6

hours and

45

minutes,

45

minutes need to be expressed in hours.

45 min convert \frac{4 5}{6 0} h = 0.75 h

Therefore,

6

hours and

45

minutes are

6.75

hours. This value can be substituted for

t

in the equation.

d = 8 t substitute d d = 8 (6.75) = 54

In

6

hours and

45

minutes, Tiffaniqua would travel

54

kilometers. Finally, to calculate how many hours it would take her to travel

44

kilometers at this speed,

44

will be substituted for

d

in the formula.

d = 8 t

Substitute

$d = 44$

44 = 8 t

DivEqn

$LHS / 8 = RHS / 8$

5.5 = t

RearrangeEqn

Rearrange equation

t = 5.5

If she kept her constant speed, it would take Tiffaniqua

5.5

hours to travel

44

kilometers.

Closure

Inverse Variation

Another type of variation is inverse variation. Here, one variable is the quotient of the constant of variation and the other variable, which cannot be zero.

y = \frac{k}{x}, x \neq = 0

Inverse variation occurs when the product of the variables is constant.

y = \frac{k}{x} \Leftrightarrow x y = k

As in direct variation, the constant of variation cannot be zero. For example, let the constant of variation of an inverse variation be

10 .

To draw its graph, a table of values will be first made. Only positive values will be considered for the

x -

variable.

$x$	$\frac{1 0}{x}$	$y = \frac{1 0}{x}$
$1$	$\frac{1 0}{1}$	$10$
$2$	$\frac{1 0}{2}$	$5$
$3$	$\frac{1 0}{3}$	$\approx 3.3$
$4$	$\frac{1 0}{4}$	$2.5$
$5$	$\frac{1 0}{5}$	$2$
$6$	$\frac{1 0}{6}$	$\approx 1.7$
$7$	$\frac{1 0}{7}$	$\approx 1.4$
$8$	$\frac{1 0}{8}$	$1.25$
$9$	$\frac{1 0}{9}$	$\approx 1.1$
$10$	$\frac{1 0}{1 0}$	$1$

Next, the points found in the table will be plotted and connected.

The graph of an inverse variation is not a straight line.

Level 1

Level 2

Level 3

Direct Variation and Proportional Relationships

Suppose

y

varies directly with

x .

How does the value of

y

change if the value of

x

is tripled?

A . B . C . D . y increases by 3 . y decreases by 3 . y is tripled . y is decreased by one third .

Since we know that y varies directly with x, we can represent the relationship with a direct variation equation. y=kx Let's say that the graph of the linear equation passes through the point ( a, b). In this case, the value of b is equal to k a. This can be shown by substituting the point ( a, b) into the equation. y=kx substitute b=k a We want to determine how the value of y changes when x is tripled. To do so, we will substitute x= 3a into the equation y=kx and find the corresponding value for y.

Now we can substitute ka= b to find y.

As we can see, when x is tripled, y is also tripled. This corresponds to option C.

Direct Variation and Proportional Relationships

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