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.
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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.
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Explore
Comparing Two Ratios
Consider different pairs of ratios. Compare their values and determine whether they are equal or not.
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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 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
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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.
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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 PropertyThe Cross Products Property can be proven by using the Properties of Equality.
a/b=c/d
MultEqn
LHS * b=RHS* b
a=c/d* b
MoveRightFacToNum
a/c* b = a* b/c
a=cb/d
MultEqn
LHS * d=RHS* d
ad=cb
CommutativePropMult
Commutative Property of Multiplication
ad=bc
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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 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.
Determine the correct solution.
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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.
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!
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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?
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b How many liters of water do they need to use for 14 of a liter of cleaning concentrate?
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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
Multiply
6=9x
DivEqn
.LHS /9.=.RHS /9.
6/9=x
RearrangeEqn
Rearrange equation
x=6/9
ReduceFrac
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
MultEqn
LHS * 4=RHS* 4
2 * y * 4=9/4 * 4
CancelCommonFac
Cancel out common factors
2 * y * 4=9/4 * 4
SimpQuot
Simplify quotient
2 * y * 4=9
Multiply
Multiply
8y=9
DivEqn
.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.
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Pop Quiz
Solving Proportions
Solve the given proportion for the unknown variable x.
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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.
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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.
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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
expand_more 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.
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
expand_more 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
expand_more Finally, plot the ordered pairs from the table of values and connect them with a straight line.
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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!
The following graph shows the cleaned area by the vacuum cleaner in x minutes.
Find the area cleaned by the robotic vacuum cleaner per minute.
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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.
This means that the vacuum cleaner can finish cleaning the living room in 25 minutes.
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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.
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Proportional and Non-proportional Relationships
Exercise 2.1
A submarine travels 21 miles in a 12 hour. How long would it take the submarine to travel 56 miles if it continues at the same rate? Round the answer to one decimal place.
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Note that the number of miles traveled and the time are proportional variables. Let's start by writing a ratio between these variables. Time/Distance Traveled=12/21 Next, since the submarine continues to travel at the same rate, we can write a proportion to represent this situation. Let x be the time needed. 12/21=x/56 Now, we will solve this proportion.
The submarine needs to travel for about 1.3 hours to complete 56 miles.
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Solution
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We want to find how many square meters can be painted by using 2 liters of paint. At first, we will find the unit rate, which is the point where (1,r) corresponds. Let's look at the graph to see this point!
The unit rate in this context refers to the area painted per one liter of paint. This means that Ali can paint 15 square meters with 1 liter of paint. In addition, these two variables are proportional because the line passes through the origin. We can now use this information to write a proportion to find the area y painted with 2 liters of paint. 15/1=y/2 We can use the Cross Products Property to find the value of y.
We got that Ali can paint 30 square meters of the wall using 2 liters of paint. We can graph this point on the given graph to see that this result matches the information in the graph.
This time we will find out the liters of paint needed for an area of 51 square meters. We can use the unit rate we got in the previous part. In this case, let x be the liters of paint needed for an area of 51 square meters. 15/1=51/x Once again, we will solve this proportion by applying the Cross Products Property.
Ali needs to use 3.4 liters of paint to paint 51 square meters.
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The graph of a proportional relationship passes through (45,27) and (1,y). Find y.
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Recall that the graph of a proportional relationship is a line and it passes through the origin. Also, since it represents a direct variation, its equation will be in the following form. y=kx In this form, k is the constant of variation. Let's rewrite the general form. y=kx ⇔ k=y/x Note that the ratio of y to x is equal to the constant of variation. With this in mind, we can find the constant of variation of our relation by using the coordinates of the point (45,27). k=y/x ⇒ k=27/45 Great! Next, recall that the y-coordinate of the point (1,y) represents the unit rate of a direct variation graph since its x-coordinate is 1. On the other hand, the unit rate and constant of variation are equivalent ratios. Knowing this leads us to write the following proportion. 27/45=y/1 Now, let's rewrite this proportion to find y.
We found that the unit rate is 35, which is the missing coordinate of the point (1,y).
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Emily and Paulina take a cycling trip together. The following graph shows how far Emily goes in miles.
The following table shows the hours and miles related to Paulina's cycling.
| Hours, x | Miles, y |
|---|---|
| 1 | 12 |
| 2 | 24 |
| 3 | 36 |
| 4 | 48 |
Who rode faster?
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We will examine the graph and the table one at a time to decide who is going faster. Let's start with the graph. Notice that the graph is a line that is passing through the origin. This means that it represents a direct variation. Let's find the point that x-coordinate is 1 in the graph.
Recall that the point (1,r) on a direct variation graph represents the unit rate. In other words, it shows how much Emily goes in one hour. Since the graph passes through the point (1, 10), Emily rides 10 miles in one hour. Now, we will examine the given table of values.
| Hours, x | Miles, y |
|---|---|
| 1 | 12 |
| 2 | 24 |
| 3 | 36 |
| 4 | 48 |
We can decide whether it represents a proportional relation between the variables x and y by checking the ratios of the numbers in each line. Let's do it!
| Hours, x | Miles, y | Mile/Hour,y/x |
|---|---|---|
| 1 | 12 | 12/1= 12 |
| 2 | 24 | 24/2= 12 |
| 3 | 36 | 36/3= 12 |
| 4 | 48 | 48/4= 12 |
The ratios in each row are equivalent, so the variables x and y are proportional. This means that the values in this table also represent a direct variation. With this in mind, we can say that Paulina goes 12 miles per hour by looking sy the values in the first line. c Emily & Paulina [0.5em] 10 miles/1 hour & 12 miles/1 hour We can conclude that Paulina rides faster than Emily.
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