Understanding Proportional and Non-Proportional Relationships: Cross Products and Direct Variation

a Start by considering the dilution rate of the cleaning concentrate.

2 : 9 = \frac{2}{9}

This means that they need to use

2

liters of cleaning concentrate for every

9

liters of water. Since she needs to keep this concentration the same, the ratio will be equivalent when she uses

3

liters of water. With this in mind, write an equivalent ratio. Let

x

be the amount of cleaning concentrate.

\frac{Cleaning Concentrate}{Water} = \frac{x}{3}

Next, set these ratios equal to get a proportion.

\frac{2}{9} = \frac{x}{3}

This proportion can be solved by using the Cross Products Property. Now, set equal the product of extremes to the product of means.

2 \cdot 3 = 9 \cdot x

Next, solve this equation for

x .

2 \cdot 3 = 9 \cdot x

Multiply

6 = 9 x

DivEqn

$LHS / 9 = RHS / 9$

\frac{6}{9} = x

RearrangeEqn

Rearrange equation

x = \frac{6}{9}

ReduceFrac

$\frac{a}{b} = \frac{a / 3}{b / 3}$

x = \frac{2}{3}

They can use

\frac{2}{3}

of a liter of cleaning concentrate for

3

liters of water.

b This time it is asked to find the amount of water to use

\frac{1}{4}

liters of cleaning concentrate. Once again, write an equivalent ratio to

2 : 9

to keep the concentration the same. Let

y

be the amount of water.

\frac{2}{9} = \frac{\frac{1}{4}}{y}

Now, apply the Cross Products Property.

2 \cdot y = 9 \cdot \frac{1}{4}

Finally, solve the obtained equation to find

y .

2 \cdot y = 9 \cdot \frac{1}{4}

MoveLeftFacToNum

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

2 \cdot y = \frac{9}{4}

MultEqn

$LHS \cdot 4 = RHS \cdot 4$

2 \cdot y \cdot 4 = \frac{9}{4} \cdot 4

CancelCommonFac

Cancel out common factors

2 \cdot y \cdot 4 = \frac{9}{4} \cdot 4

SimpQuot

Simplify quotient

2 \cdot y \cdot 4 = 9

Multiply

8 y = 9

DivEqn

$LHS / 8 = RHS / 8$

y = \frac{9}{8}

This means that they need to dilute

\frac{9}{8}

of a liter of water to use

\frac{1}{4}

of cleaning concentrate.

$x$	$y = 3 x$	$y$	$(x, y)$
$- 1$	$y = 3 (- 1)$	$- 3$	$(- 1, - 3)$
$0$	$y = 3 (0)$	$0$	$(0, 0)$
$1$	$y = 3 (1)$	$3$	$(1, 3)$
$2$	$y = 3 (2)$	$6$	$(2, 6)$
$3$	$y = 3 (3)$	$9$	$(3, 9)$

x

y = 3 x

y

(x, y)

- 1

y = 3 (- 1)

- 3

(- 1, - 3)

0

y = 3 (0)

0

(0, 0)

1

y = 3 (1)

3

(1, 3)

2

y = 3 (2)

6

(2, 6)

3

y = 3 (3)

9

(3, 9)

$x$	$y = k x$	$y$
$- 2$	$y = 0 \cdot (- 2)$	$0$
$- 1$	$y = 0 \cdot (- 1)$	$0$
$1$	$y = 0 \cdot 1$	$0$
$2$	$y = 0 \cdot 2$	$0$

x

y = k x

y

- 2

y = 0 \cdot (- 2)

0

- 1

y = 0 \cdot (- 1)

0

1

y = 0 \cdot 1

0

2

y = 0 \cdot 2

0

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Catch-Up and Review

Comparing Two Ratios

The Concept of Proportion

Proportion

Finding Equal Ratios

Cross Products Property

Proof

Learning to Apply Cross Products Property Correctly

Hint

Solution

Dylan's Solution

Mark's Solution

Correct Solution

Dilution of a Cleaning Concentrate

Hint

Solution

Solving Proportions

Direct Variation

Identifying the Constant of Variation

Graphing a Direct Variation Using a Table of Values

Perfomance of a Robotic Vacuum Cleaner

Hint

Solution

Can the Constant of Variation of a Direct Variation Be Zero?

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