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5-6 Direct Variation

Lesson QuizLesson Quiz

Lesson PresentationLesson Presentation

Warm UpWarm Up


Warm UpSolve for y.1. 3 + y = 2x 2. 6x = 3y

Write an equation that describes the relationship.

3.

y = 2xy = 2x – 3

4. 5.

y = 3x

9 0.5

Solve for x.


MA.912.A.3.12 Graph a linear equation…in two variables…Also MA.912.A.2.1, MA.912.A.2.13.

Sunshine State Standards


Identify, write, and graph direct variation.

Objective


Vocabulary

direct variationconstant of variation


A recipe for paella calls for 1 cup of rice to make 5 servings. In other words, a chef needs 1 cup of rice for every 5 servings.

The equation y = 5x describes this relationship. In this relationship, the number of servings varies directly with the number of cups of rice.


A direct variation is a special type of linear relationship that can be written in the form y = kx, where k is a nonzero constant called the constant of variation.


Additional Example 1A: Identifying Direct Variations from Equations

Tell whether the equation represents a direct variation. If so, identify the constant of variation. y = 3x

This equation represents a direct variation because it is in the form of y = kx. The constant of variation is 3.


3x + y = 8 Solve the equation for y.Since 3x is added to y, subtract 3x

from both sides.–3x –3x

y = –3x + 8

This equation is not a direct variation because it cannot be written in the form y = kx.

Additional Example 1B: Identifying Direct Variations from Equations

Tell whether the equation represents a direct variation. If so, identify the constant of variation.


–4x + 3y = 0 Solve the equation for y.Since –4x is added to 3y, add 4x

to both sides.

+4x +4x3y = 4x

This equation represents a direct variation because it is in the form of y = kx. The constant of variation is .

Since y is multiplied by 3, divide both sides by 3.

Additional Example 1C: Identifying Direct Variations from Equations



Check It Out! Example 1a

3y = 4x + 1

This equation is not a direct variation because it is not written in the form y = kx.



Check It Out! Example 1c

y + 3x = 0 Solve the equation for y.Since 3x is added to y, subtract 3x

from both sides.

– 3x –3xy = –3x

This equation represents a direct variation because it is in the form of y = kx. The constant of variation is –3.



What happens if you solve y = kx for k?

y = kx

So, in a direct variation, the ratio is equal to the constant of variation. Another way to identify a direct variation is to check whether is the same for each ordered pair (except where x = 0).

Divide both sides by x (x ≠ 0).


Additional Example 2A: Identifying Direct Variations from Ordered Pairs

Tell whether the relationship is a direct variation. Explain.

Method 1 Write an equation.

y = 3x

This is direct variation because it can be written as y = kx, where k = 3.

Each y-value is 3 times the corresponding x-value.


Additional Example 2A Continued


Method 2 Find for each ordered pair.

This is a direct variation because is the same for each ordered pair.



y = x – 3 Each y-value is 3 less than the corresponding x-value.

This is not a direct variation because it cannot be written as y = kx.

Additional Example 2B: Identifying Direct Variations from Ordered Pairs




This is not direct variation because is the not the same for all ordered pairs.

Additional Example 2B Continued


…



Check It Out! Example 2b


y = –4x Each y-value is –4 times the corresponding x-value .

This is a direct variation because it can be written as y = kx, where k = –4.



Check It Out! Example 2c


This is not direct variation because is the not the same for all ordered pairs.


If you know one ordered pair that satisfies a direct variation, you can write the equation. You can also find other ordered pairs that satisfy the direct variation.


Additional Example 3: Writing and Solving Direct Variation Equations

The value of y varies directly with x, and y = 3, when x = 9. Find y when x = 21.

Method 1 Find the value of k and then write the equation.

y = kx Write the equation for a direct variation.

3 = k(9) Substitute 3 for y and 9 for x. Solve for k.

Since k is multiplied by 9, divide both sides by 9.

The equation is y = x. When x = 21, y = (21) = 7.


The value of y varies directly with x, and y = 3 when x = 9. Find y when x = 21.

Method 2 Use a proportion.

9y = 63

y = 7

In a direct variation is the same for all values of x and y.

Use cross products.

Since y is multiplied by 9 divide both sides by 9.

Additional Example 3 Continued


Check It Out! Example 3

The value of y varies directly with x, and y = 4.5 when x = 0.5. Find y when x = 10.

Method 1 Find the value of k and then write the equation.

y = kx Write the equation for a direct variation.

4.5 = k(0.5) Substitute 4.5 for y and 0.5 for x. Solve for k.

Since k is multiplied by 0.5, divide both sides by 0.5.

The equation is y = 9x. When x = 10, y = 9(10) = 90.

9 = k


Check It Out! Example 3 Continued

Method 2 Use a proportion.

0.5y = 45

y = 90

In a direct variation is the same for all values of x and y.

Use cross products.

Since y is multiplied by 0.5 divide both sides by 0.5.

The value of y varies directly with x, and y = 4.5 when x = 0.5. Find y when x = 10.


Check It Out! Example 4

The perimeter y of a square varies directly with its side length x. Write a direct variation equation for this relationship. Then graph.

Step 1 Write a direct variation equation.

perimeter = 4 sides times length

y = 4 • x



Step 2 Choose values of x and generate ordered pairs.

x y = 4x (x, y)

0 y = 4(0) = 0 (0, 0)

1 y = 4(1) = 4 (1, 4)

2 y = 4(2) = 8 (2, 8)



Step 3 Graph the points and connect.




Standard Lesson Quiz

Lesson Quizzes

Lesson Quiz for Student Response Systems


Lesson Quiz: Part ITell whether each equation represents a direct variation. If so, identify the constant of variation.

1. 2y = 6x yes; 3

2. 3x = 4y – 7 no


3.


Lesson Quiz: Part II

4. The value of y varies directly with x, and y = –8 when x = 20. Find y when x = –4. 1.6

5. Apples cost $0.80 per pound. Write a direct variation equation to describe the cost y of x pounds of apples. Then graph.

2

4

6

y = 0.8x



1. Identify the equation that represents

direct variation and its constant of variation.

A. 4y = 12x; 3

B. 4x = 3y − 8; 2

C. 4y = 5x + 3; 5

D. 3y = 5x + 11; 3



2. Identify the equation that does not

represent direct variation.

A. 6x = 8y − 14

B. 6y = 12x

C. 18x = 3y

D. 4y = 12x

5-6 Direct VariationLesson Quiz for Student Response Systems

3. Identify the relationship that represents a direct variation.

x 4 28 40

y 15 7 10

x 15 19 23

y 5 8 11

x 6 4 2

y 5 7 9

x 2 4 8

y 10 20 40A.

B.

C.

D.

5-6 Direct VariationLesson Quiz for Student Response Systems

4. Identify the relationship that does not represent a direct variation.

x 5 7 9

y 1 2 3

x 6 12 18

y 3 6 9

x 2 4 8

y 10 20 40

x 4 8 12

y 10 20 30A.

B.

C.

D.



5. The value of y varies directly with x, and

y = –5 when x = 25. Identify the value of y

when x = –10.

A. y = –1

B. y = 0

C. y = 2

D. y = –2