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Pre-Algebra View details

5. Proportional and Non-proportional Relationships

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Chapter 5

5.

Proportional and Non-proportional Relationships

This lesson explores the mathematical concepts of proportional and non-proportional relationships. It discusses how to identify these relationships through equations and graphs. The Cross Products Property is highlighted as a method to prove whether two ratios are proportional. Direct variation is another focus, showing how a constant rate of change between two variables can indicate a proportional relationship. Practical examples include calculating the area cleaned by a robotic vacuum cleaner and comparing the cost-effectiveness of buying apple juice from different stores. This information is crucial for anyone needing to understand how variables relate to each other, whether in academics, finance, or engineering.

Lesson Settings & Tools

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

Proportional and Non-proportional Relationships

Slide of 13

Certain situations need rates to be constant. For instance, playing an arcade game hourly requires paying more money at a constant rate. Such a case is said to have a proportional relationship between the given quantities. There are also cases of non-porportional relationships. This lesson will introduce various ones.

Catch-Up and Review

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

Ratio
Fraction
Rational Numbers
Variable
Constant

Explore

Comparing Two Ratios

Consider different pairs of ratios. Compare their values and determine whether they are equal or not.

Comparing two different ratios

If the ratios are equal, how can this fact be written algebraically using only mathematics symbols?

Discussion

The Concept of Proportion

The value of two equivalent ratios is the same. Because of this, an equals sign can be written between two equivalent ratios to create a proportion.

Concept

Proportion

A proportion is an equation showing the equivalence of two ratios, or fractions, with different numerators and denominators. a/b = c/d or a:b=c:d The first and last numbers in the proportion are called the extremes, while the other two numbers are called the means. ↓ a0.75em extremes means : ↑ b= ↑ c:↓ d

As an example of proportionality, consider slices of pizza. Depending on the number of times it has been sliced, the same amount of pizza could be cut into 1, 2, or 4 pieces.

In this case, one-third of a pizza is the same amount of pizza as two-sixths or four-twelfths. If the simplified forms of two fractions are equal, then they are said to be proportional. For example, one-third is proportional to two-sixths and four-twelfths.

In contrast, two ratios have a non-proportional relationship if they are not equivalent. Consider, for example, 23 and 32. These ratios are in their simplest form, but they are not equal. This means that they are non-proportional. This relationship is shown by writing an inequality symbol between the ratios.

2/3 ≠ 3/2

Pop Quiz

Finding Equal Ratios

Consider the given ratio. Then, analyze the values of the ratios in each answer and choose which one forms a proportion with the given ratio.

Random generator that shows four ratios

Discussion

Cross Products Property

In a proportion, the product of the extremes is equal to the product of the means.

a/b=c/d ⇒ ad=bc

This property is also known as cross-multiplication or Means-Extremes Property of Proportion.

Proof

Cross Products Property

The Cross Products Property can be proven by using the Properties of Equality.

a/b=c/d

LHS * b=RHS* b

a=c/d* b

MoveRightFacToNum

a/c* b = a* b/c

a=cb/d

LHS * d=RHS* d

ad=cb

CommutativePropMult

Commutative Property of Multiplication

ad=bc

Example

Learning to Apply Cross Products Property Correctly

Dylan's mom is making chocolate chip cookies. She uses a recipe which claims that for every three cups of flour, two cups of white sugar must be added. However, she wants to make a larger batch so she uses six cups of white sugar for nine cups of flour. She is unsure if she is using the correct amount.

Dylan's mom's recipe to make chocolate chips

Dylan and his brother Mark help their mother by forming a proportion with the given white sugar and flour amounts to check whether their mother has the same rate of ingredients. Although both of the guys used the Cross Products Property, their solutions were different.

Dylan's and Mark's solutions

Determine the correct solution.

Hint

Recall what the Cross Products Property states. Then identify the extremes and means in the proportion set by the boys. Are the second steps in either solution correct?

Solution

Begin by analyzing each solution separately.

Dylan's Solution

Dylan began by equating the given ratios to investigate if they form a proportion. He put a question mark above the equals sign until he knows if a proportion is formed or not.

Focus on where Dylan applied the Cross Products Property.

Recall what that property states to determine whether it was correctly applied.

a/b=c/d ⇒ ad= bc

The property claims that the product of the extremes is equal to the product of the means. Identify the extremes and means in the proportion written by Dylan.

Note that the means and the extremes should be multiplied and set equal to apply the property correctly. However, Dylan mistakenly multiplied the numerator and denominator of each fraction. 2* 3 &= 6* 9 * 2* 9 &= 3* 6 ✓ This means that Dylan's solution is not correct.

Mark's Solution

Next, Mark's solution will be analyzed. After writing the ratios as a likely proportion, he also chose to apply the Cross Products Property.

Notice that Mark multiplied the numerators and denominators of the fractions and then set them equal instead of multiplying the extremes and means of the fractions. 2* 6 &= 3* 9 * 2* 9 &= 3* 6 ✓ Mark's solution is also incorrect. In other words, none of the given solutions are correct.

Correct Solution

Now, it is time to solve the exercise properly.

2/3? =6/9

Cross multiply

2* 9 ? =3* 6

Multiply

18 = 18

Since the cross products are equal, the ratios form a proportion. This is great news. Their mom already chose the ingredients in the correct amounts. However, the boys made the mistakes in their calcuations and the cookies came out tasting nasty!

Example

Dilution of a Cleaning Concentrate

The family made a huge mess while baking the cookies. They need to clean the kitchen. Some instructions state that the dilution rate for cleaning concentrate is 2:9. This means 9 liters of water is used to 2 liters of cleaning concentrate.

a How many liters of cleaning concentrate do they need to use for 3 liters of water?

b How many liters of water do they need to use for 14 of a liter of cleaning concentrate?

Hint

a Write the corresponding proportion and use the Cross Products Property to find the missing value in the proportion.

b Write the corresponding proportion and use the Cross Products Property to find the missing value in the proportion.

Solution

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

2:9=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. Cleaning Concentrate/Water=x/3 Next, set these ratios equal to get a proportion. 2/9=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* 3=9 * x Next, solve this equation for x.

2* 3=9 * x

Multiply

6=9x

.LHS /9.=.RHS /9.

6/9=x

Rearrange equation

x=6/9

a/b=.a /3./.b /3.

x=2/3

They can use 23 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 14 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.

2/9=14/y Now, apply the Cross Products Property. 2 * y=9 * 1/4 Finally, solve the obtained equation to find y.

2 * y=9 * 1/4

MoveLeftFacToNum

a*b/c= a* b/c

2 * y=9/4

LHS * 4=RHS* 4

2 * y * 4=9/4 * 4

CancelCommonFac

Cancel out common factors

2 * y * 4=9/4 * 4

Simplify quotient

2 * y * 4=9

Multiply

8y=9

.LHS /8.=.RHS /8.

y=9/8

This means that they need to dilute 98 of a liter of water to use 14 of cleaning concentrate.

Pop Quiz

Solving Proportions

Solve the given proportion for the unknown variable x.

Different proportions with variable x are generated randomly.

Discussion

Direct Variation

Direct variation is a relationship between two variables, x and y, where an increase in one variable causes the other to increase by a constant factor k. This means that if x increases, y increases, and if x decreases, y decreases. The following equation shows this relationship algebraically.

y=kx

The constant k is the constant of variation or proportionality, which connects x and y. In other words, k shows the amount of change in y for each unit increase in x. The constant of variation can be any real number except 0. The constant of variation can be found by dividing y by x. y=kx ⇔ y/x=k An example of direct variation is the connection between the number of hours worked and the money earned. If the constant of proportionality is $10 per hour, this relationship can be expressed with following equation. Money Earned = $10 * Hours Worked The money earned by someone who worked 5 hours can be found by multiplying $10 by 5. Money Earned = $10 * 5 = $50 In addition, the amount earned by someone who worked 8 hours can be calculated using a similar process. Money Earned = $10 * 8 = $80

Notice that the amount earned increases as the number of hours worked increases, and the rate of increase is constant at $10 per hour. This is the essence of direct variation.

Pop Quiz

Identifying the Constant of Variation

Recall that direct variation equations can be written in the forms y=kx, k= yx, or x= yk. Identify the constant of variation k in the given direct variation equation. If the value of k is a fraction, write it in its simplest form.

Identifying the Constant of Variation

Discussion

Graphing a Direct Variation Using a Table of Values

A direct variation can be represented graphically by a line that passes through the origin on a coordinate plane. A table of values can help get the line of a direct variation given in the form y=kx. Consider, for example, the following equation of a direct variation. y=3x Three steps can be followed to draw the line for this equation.

1

Write the Equation in the form y=kx

Recall that a direct variation equation can be written in the following forms. y=kx, y/x=k, or y/k=x If a direct variation equation is in the form of yx=k or yk=x, then rewrite it to be in the form of y=kx. This form will help to apply the next steps more straightforwardly.

y/x=k

LHS * x=RHS* x

y=kx

Note that the given equation is already in the form y=kx where the constant of variation k is 3.

2

Make a Table of Values

A table of values can be made by substituting several random values for x and then solving the equation for y.

x	y=3x	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)

3

Plot the Ordered Pairs and Graph the Line

Finally, plot the ordered pairs from the table of values and connect them with a straight line.

Graph of the equation y=3*x with points (-1, -3), (0, 0), (1, 3), (2, 6), and (3, 9).

It is worth to keep in mind that if a line does not pass through the origin then it does not represent a direct variation. In other words, there is not a proportional relationship between the variables x and y.

Example

Perfomance of a Robotic Vacuum Cleaner

They will use a robotic vacuum cleaner to clean the floor of not only their kitchen but their entire house!

Robotic Vacuum Cleaner

The following graph shows the cleaned area by the vacuum cleaner in x minutes.

The diagram shows the line y=10x on the first quadrant. x represents time in minutes, and y represents cleaned area in square feet. x-axis ranges from 0 to 7 with increments of 1, and y-axis ranges from 0 to 25 with increments of 5.

Find the area cleaned by the robotic vacuum cleaner per minute.

If the living room is 250 square feet, how many minutes does it take for the robotic vacuum cleaner to finish cleaning this room?

Hint

The graph of a proportional relationship is a line that passes through the origin. This direct variation can be represented by an equation in the form y=kx.

Solution

Notice that the given graph is a line that passes through the origin. This means that it represents a proportional relationship between the variables. With this in mind start by examining several points on the graph.

Recall that a point (x,y) in this graph represents the cleaned area y in x minutes. Now take a look at the point which x-coordinate is 1. (1, 10) The y-coordinate of (1, 10) is 10. This means that the robotic vacuum cleaner can clean 10 square feet in one minute. For a point as (1,r) in a direct variation graph, r represents the unit rate. In the context of the problem, unit rate means the area cleaned in one minute or per minute by the vacuum cleaner. (1, r) ⇒ r=Unit Rate Now, remember the general form of a direct variation equation. y=kx In this form, k is the constant of variation or the unit rate. With this in mind, substitute k=10 into this equation to have an equation representing the cleaned area by the robotic vacuum cleaner according to the time in minutes. y=10x Next, use this equation to find the time for cleaning the 250 square feet area. Notice that 250 represents the y-coordinate and it is asked to find the corresponding x-coordinate at that point. Now, substitute y=250 into the equation and solve it for x.

y=10x

y= 250

250=10x

.LHS /10.=.RHS /10.

250/10=10x/10

Calculate quotient

25=x

This means that the vacuum cleaner can finish cleaning the living room in 25 minutes.

Closure

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

Recall that if the variables x and y are proportional, then there exists a direct variation between them. This relation can be represented by the following equation. y=kx The value of k is the constant of variation. What if k=0? Make a table of values to see what happens when k is equal to zero.

x	y=kx	y
-2	y=0 * (-2)	0
-1	y=0 * (-1)	0
1	y=0 * 1	0
2	y=0 * 2	0

As it can be seen from the table, all the y values become 0 for any value of x. Since the variable y stays the same while the variable x is changing, it does not exist a direct variation between the variables. When k=0, its equation becomes y=0. y=0 * x ⇒ y=0 Note that its graph is also a line passing through the origin.

line of y=0

This means that if a line passes through the origin, it cannot necessarily be said that it is the graph of a direct variation. It must also be known that the constant of variation is not equal to zero.

Proportional and Non-proportional Relationships

Exercise 3.1

Consider the following two proportions. a/b=3/4 and b/c=2/7 What is the ratio ac?

We are given two different proportions such that they share the same variable b. a/b=3/4 and b/c=2/7 The corresponding values of b are different in both proportions. The value of b in the first proportion is twice that of b in the second. We can create an equivalent ratio to the second ratio by expanding that ratio by 2. That can help us to have the same value for b on both proportions. b/c=2* 2/7* 2 ⇒ b/c=4/14 Now, both of the b values are represented by 4. This leads us to form a relation between the variables a and c. Since the same number represents the common variable b, the remaining variables are now proportional. a/b=3/4 and b/c=4/14 Let's use 3 for a and 14 for c to write a ratio for ac. a/c=3/14 Notice that 314 is already in its simplest form.

Proportional and Non-proportional Relationships

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