grade 8 algebra1 direct variation

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CONFIDENTIAL 1 Grade 8 Algebra1 Grade 8 Algebra1 Direct Variation Direct Variation

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Grade 8 Algebra1 Direct Variation. Warm Up. Find the slope of the line described by each equation. 1) 4x + y = -9. 2) 6x - 3y = -9. 3) 5x = 10y - 5. Direct Variation. - PowerPoint PPT Presentation

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Page 1: Grade 8 Algebra1 Direct Variation

CONFIDENTIAL 1

Grade 8 Algebra1Grade 8 Algebra1

Direct VariationDirect Variation

Page 2: Grade 8 Algebra1 Direct Variation

CONFIDENTIAL 2

Warm UpWarm UpFind the slope of the line described by

each equation.

1) 4x + y = -9

2) 6x - 3y = -9

3) 5x = 10y - 5

Page 3: Grade 8 Algebra1 Direct Variation

CONFIDENTIAL 3

Direct Variation

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 .

A recipe for chili calls for 2 pounds of beef to make 4 servings. In other words, a chef needs 2 pounds for every 4 servings.

The equation y = 2x describes this relationship. In this relationship, the number of servings varies directly with the

number of pounds of beef.

Beef (lb) x 2 3 4 5

Servings y 4 6 8 10

Page 4: Grade 8 Algebra1 Direct Variation

CONFIDENTIAL 4

1) y = 4x

Identifying Direct Variations from Equations

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

This equation represents a direct variation because it is in the form y = kx. The constant of variation is 4.

Next Slide

Page 5: Grade 8 Algebra1 Direct Variation

CONFIDENTIAL 5

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

5

Since y is multiplied by 5, divide both sides by 5.

5y = 3x 5 5

y = 3x 5

Solve the equation for y.

2) -3x + 5y = 0

-3x + 5y = 0

+3x +3x

5y = 3x

Since -3x is added to y, add 3x to both sides.

Next Slide

Page 6: Grade 8 Algebra1 Direct Variation

CONFIDENTIAL 6

Solve the equation for y.

Since 2x is added to y, subtract 2x from both sides.y =-2x + 10

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

3) 2x + y = 10

2x + y = 10

-2x -2x

Page 7: Grade 8 Algebra1 Direct Variation

CONFIDENTIAL 7

Now you try!

1a) 3y = 4x + 1

1b) 3x = -4y

1c) y + 3x = 0

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

Page 8: Grade 8 Algebra1 Direct Variation

CONFIDENTIAL 8

What happens if you solve y = kx for k?

y = kxx x

y = kx

So, in a direct variation, the ratio y is equal to the x

constant of variation.

Another way to identify a direct variation is to check

whether y is the same for each ordered pair

(except where x = 0).

x

Divide both sides by x (x ≠ 0)

y = kx

Divide both sides by x (x ≠ 0)

Page 9: Grade 8 Algebra1 Direct Variation

CONFIDENTIAL 9

Identifying Direct Variations from Ordered Pairs

Tell whether each relationship is a direct variation. Explain.

x 1 3 5

y 6 18 30

1)

Method 1: Write an equation.

Each y-value is 6 times the corresponding x-value.y = 6x

This is a direct variation because it can be written as y = kx, where k = 6.

6 = 61

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

Method 2: Find y for each ordered pair.x

18 = 6 3

30 = 6 5

Page 10: Grade 8 Algebra1 Direct Variation

CONFIDENTIAL 10

x 2 4 8

y -2 0 4

2)

Method 1: Write an equation.

Each y-value is 4 less than the corresponding x-value.y = x - 4

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

-2 = -1 2

This is not a direct variation because y is not the same xfor each ordered pair.

Method 2: Find y for each ordered pair.x

0 = 04

4 = 18 2

Page 11: Grade 8 Algebra1 Direct Variation

CONFIDENTIAL 11

Now you try!

2c)

Tell whether each relationship is a direct variation. Explain.

2a) 2b)

Page 12: Grade 8 Algebra1 Direct Variation

CONFIDENTIAL 12

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.

Page 13: Grade 8 Algebra1 Direct Variation

CONFIDENTIAL 13

Writing and Solving Direct Variation Equations

The value of y varies directly with x, and y = 6 when x = 12.Find y when x = 27.

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

Write the equation for a direct variation.y = kx

The equation is y = 1 x . When x = 27, y = 1 (27) = 13.5. 2 2

Substitute 6 for y and 12 for x. Solve for k.6 = k(12)

Since k is multiplied by 12, divide both sides by 12.6 = k(12)

Page 14: Grade 8 Algebra1 Direct Variation

CONFIDENTIAL 14

Method 2: Use a proportion.

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

6 = y12 27

Use cross products.12y = 162

Since y is multiplied by 12, divide both sides by 12.y = 13.5

Page 15: Grade 8 Algebra1 Direct Variation

CONFIDENTIAL 15

Now you try!

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

Page 16: Grade 8 Algebra1 Direct Variation

CONFIDENTIAL 16

Graphing Direct Variations

1) The three-toed sloth is an extremely slow animal. On the ground, it travels at a speed of about 6 feet per

minute. Write a direct variation equation for the distance y a sloth will travel in x minutes. Then graph.

Step1: Write a direct variation equation.

distance = 6 feet per minute

Step2: Choose values of x and generate ordered pairs.

X y = 6x (x, y)

0 y = 6 (0) = 0 (0, 0)

1 y = 6 (1) = 6 (1, 6)

2 y = 6 (2) = 12 (2, 12)

Page 17: Grade 8 Algebra1 Direct Variation

CONFIDENTIAL 17

Step3: Graph the points and connect.

Look at the graph. It passes through (0, 0) and has a slope of 6. The graph of any direct variation y = kx

• is a line through (0, 0) • has a slope of k.

Page 18: Grade 8 Algebra1 Direct Variation

CONFIDENTIAL 18

Now you try!

4) The perimeter y of a square varies directly with its side length x. Write a direct variation equation

for this relationship. Then graph.

Page 19: Grade 8 Algebra1 Direct Variation

CONFIDENTIAL 19

BREAK

Page 20: Grade 8 Algebra1 Direct Variation
Page 21: Grade 8 Algebra1 Direct Variation

CONFIDENTIAL 21

Assessment

1) y = 4x + 9

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

2) 2y = -8x

3) x + y = 0

Page 22: Grade 8 Algebra1 Direct Variation

CONFIDENTIAL 22

Tell whether each relationship is a direct variation. Explain.

5)

4)

Page 23: Grade 8 Algebra1 Direct Variation

CONFIDENTIAL 23

6) The value of y varies directly with x, and y = -3 when x = 1. Find y when x = -6.

7) The value of y varies directly with x, and y = 6 when x = 18. Find y when x = 12.

Page 24: Grade 8 Algebra1 Direct Variation

CONFIDENTIAL 24

8) Cameron earns $5 per hour at her after-school job. The total amount of her paycheck varies directly with the amount of time she works. Write a direct variation equation for the amount of money y that she earns for working x hours. Then graph.

Page 25: Grade 8 Algebra1 Direct Variation

CONFIDENTIAL 25

9) (2, 10)

10) (-3, 9)

Each ordered pair is a solution of a direct variation. Write the equation of direct variation. Then graph

your equation and show that the slope of the line is equal to the constant of variation.

Page 26: Grade 8 Algebra1 Direct Variation

CONFIDENTIAL 26

Direct Variation

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 .

A recipe for chili calls for 2 pounds of beef to make 4 servings. In other words, a chef needs 2 pounds for every 4 servings.

The equation y = 2x describes this relationship. In this relationship, the number of servings varies directly with the

number of pounds of beef.

Beef (lb) x 2 3 4 5

Servings y 4 6 8 10

Let’s review

Page 27: Grade 8 Algebra1 Direct Variation

CONFIDENTIAL 27

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

5

Since y is multiplied by 5, divide both sides by 5.

5y = 3x 5 5

y = 3x 5

Solve the equation for y.

2) -3x + 5y = 0

-3x + 5y = 0

+3x +3x

5y = 3x

Since -3x is added to y, add 3x to both sides.

Next Slide

Identifying Direct Variations from Equations

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

Page 28: Grade 8 Algebra1 Direct Variation

CONFIDENTIAL 28

Solve the equation for y.

Since 2x is added to y, subtract 2x from both sides.y =-2x + 10

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

3) 2x + y = 10

2x + y = 10

-2x -2x

Page 29: Grade 8 Algebra1 Direct Variation

CONFIDENTIAL 29

What happens if you solve y = kx for k?

y = kxx x

y = kx

So, in a direct variation, the ratio y is equal to the x

constant of variation.

Another way to identify a direct variation is to check

whether y is the same for each ordered pair

(except where x = 0).

x

Divide both sides by x (x ≠ 0)

y = kx

Divide both sides by x (x ≠ 0)

Page 30: Grade 8 Algebra1 Direct Variation

CONFIDENTIAL 30

Identifying Direct Variations from Ordered Pairs

Tell whether each relationship is a direct variation. Explain.

x 1 3 5

y 6 18 30

1)

Method 1: Write an equation.

Each y-value is 6 times the corresponding x-value.y = 6x

This is a direct variation because it can be written as y = kx, where k = 6.

6 = 61

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

Method 2: Find y for each ordered pair.x

18 = 6 3

30 = 6 5

Page 31: Grade 8 Algebra1 Direct Variation

CONFIDENTIAL 31

Writing and Solving Direct Variation Equations

The value of y varies directly with x, and y = 6 when x = 12.Find y when x = 27.

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

Write the equation for a direct variation.y = kx

The equation is y = 1 x . When x = 27, y = 1 (27) = 13.5. 2 2

Substitute 6 for y and 12 for x. Solve for k.6 = k(12)

Since k is multiplied by 12, divide both sides by 12.6 = k(12)

Page 32: Grade 8 Algebra1 Direct Variation

CONFIDENTIAL 32

Method 2: Use a proportion.

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

6 = y12 27

Use cross products.12y = 162

Since y is multiplied by 12, divide both sides by 12.y = 13.5

Page 33: Grade 8 Algebra1 Direct Variation

CONFIDENTIAL 33

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