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Illustrative Mathematics (6-8 OUR Mathematics)

6th Grade Resources

U1 , U2 , U3 , U4 , U5 , U6 , U7 , U8

7th Grade Resources

8th Grade Resources

Preparation for High School Algebra or Integrated Math 1

University High School Summer Bridge Course

High School Illustrative M athematics - McGra w Hill, Kendall Hunt, Imagine Learning, Learn Zillion)

Algebra 1 Resources

U1 , U2, U3, U4, U5, U6, U7

OPEN UP RESOURCES Mathematics - OUR

OUR - Algebra 1 Resources

U1 , U2 , U3, U4, U5, U6, U7, U8, U9

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Master Math 7 with Morgan: Get Help with Homework!

Updated on Dec 26,2023

Table of Contents

• Introduction
• Using Equations to Solve Problems 2.1 Organizing and Ordering Values 2.2 Dividing Numbers 2.3 Solving Equations for Price 2.4 Solving Equations for Number of Tickets Sold
• Recycling Aluminum Cans 3.1 Weight and Value of Aluminum Cans 3.2 Proportional Relationship and Unit Rate 3.3 Solving Equations for Number of Cans and Value
• Proportional Relationships in Distance and Speed 4.1 Equation Relating Distance and Speed 4.2 Solving for Distance and Time
• Proportional Relationships in Volume and Bottles 5.1 Equation Relating Bottles and Volume 5.2 Solving for Volume and Number of Bottles
• Conversion of Kilometers to Miles 6.1 Proportional Relationship between Kilometers and Miles 6.2 Writing Equations and Solving for Distance

Proportional Relationship between Quarters and Toonies

Completing Tables and Identifying Constants of Proportionality

Determining Scaled Copies of Polygons

Using Equations to Solve Problems

In this lesson, we will explore the use of equations to solve various problems. We will begin by organizing and ordering values and then move on to dividing numbers. Next, we will learn how to solve equations for price and the number of tickets sold.

Organizing and Ordering Values

To solve problems efficiently, it is important to organize and order values correctly. By arranging values from least to greatest, we can effectively analyze the data and make accurate calculations.

For example, let's consider a Scenario where We Are dividing a number by different values, such as 42.6, 42.646, and 426. By moving the decimal point one place to the left, we can convert 426 to 42.6. This allows us to compare and analyze the results, determining the smallest and largest divisor and their corresponding decimal values.

Dividing Numbers

When dividing a number by various divisors, it is crucial to understand that the larger the divisor, the smaller the result will be. By assigning variables (a, b, c, and d) to represent the values, we can analyze and estimate the results.

For example, if we consider the values 42.6 divided by 7, the result is approximately 6.08571. In comparison, if we divide 42.6 by 426, the result is approximately 0.1. By estimating and rounding, we can quickly determine the approximate values of the division.

Solving Equations for Price

In certain situations, we may need to determine the price for a single ticket Based on the total sales and the number of tickets sold. By setting up an equation, we can easily solve for the unknown variable.

For instance, let's consider a scenario where a performer aims to sell 5,000 tickets and make a total of $311,000 in sales. Assuming all tickets have the same price (constant proportionality), we can set up the equation: Total Sales = Price per Ticket × Number of Tickets Sold.

By dividing the total sales ($311,000) by the number of tickets sold (5,000), we find that the price per ticket is $62.20. Therefore, each ticket should cost $62.20 for the performer to achieve their sales goal.

Solving Equations for Number of Tickets Sold

Conversely, we may also need to determine the number of tickets sold based on the total sales and the price per ticket. By rearranging the equation, we can find the value of the unknown variable.

Suppose the total sales amount is $379,420 and the price per ticket is $62.20. We can set up the equation: Total Sales = Price per Ticket × Number of Tickets Sold.

By dividing both sides of the equation by $62.20, we find that the performer sold approximately 6,100 tickets. Thus, the number of tickets sold can be calculated by dividing the total sales by the price per ticket.

Recycling Aluminum Cans

In this section, we will explore the concept of recycling aluminum cans and its related equations. We will discuss the weight and value of aluminum cans, as well as their proportional relationship and unit rates.

Weight and Value of Aluminum Cans

Aluminum cans can be recycled instead of being thrown in the garbage. Each can typically weighs 0.16 kilograms and has a value of $0.14.

To determine the weight and value of multiple cans, we can set up a table. For example, if we have 10 cans, the total weight would be 1.6 kilograms, and the value would be $1.40.

Proportional Relationship and Unit Rate

When analyzing the weight and value of aluminum cans, we observe a proportional relationship. The weight and the value of the cans have a constant proportionality, which can be represented by the equation: Value = Constant Proportionality × Weight.

By dividing the value ($0.14) by the weight (0.16 kilograms), we find that the constant proportionality (unit rate) is 0.875. This means that for every kilogram, the cans have a value of $0.875.

Solving Equations for Number of Cans and Value

To find the number of cans or the value based on a given weight or vice versa, we can use the equation: Number of Cans or Value = Constant Proportionality × Weight.

For example, if we want to determine the number of cans for a weight of 2.4 kilograms, we plug in the values into the equation: Number of Cans = Constant Proportionality × Weight. By substituting the constant proportionality (0.875) and the weight (2.4 kilograms), we find that the number of cans is approximately 150.

Similarly, if we want to find the value of 51 cans, we use the equation: Value = Constant Proportionality × Number of Cans. By substituting the constant proportionality (0.014) and the number of cans (51), we find that the value is $0.714.

Proportional Relationships in Distance and Speed

In this section, we will explore the concept of proportional relationships in distance and speed. We will discuss the equation relating distance and speed and how it can be used to solve for distance and time.

Equation Relating Distance and Speed

When a car travels at a constant speed, the distance it travels can be represented by the equation: Distance = Speed × Time. In this equation, distance is measured in miles, speed is measured in miles per hour (mph), and time is measured in hours.

For example, if a car is traveling at a speed of 65 mph for 1.5 hours, we can calculate the distance it travels by multiplying the speed and time: Distance = 65 mph × 1.5 hours = 97.5 miles.

Solving for Distance and Time

Using the equation Distance = Speed × Time, we can solve for distance or time if two of the variables are known. By rearranging the equation, we can isolate the variable we want to solve for.

For instance, if we know the distance is 26 miles and the speed is 65 mph, we can solve for the time it takes to travel that distance. Rearranging the equation as Time = Distance / Speed, we find that the time is 0.4 hours or 24 minutes.

Proportional Relationships in Volume and Bottles

In this section, we will explore the concept of proportional relationships in volume and bottles. We will discuss the equation relating the number of bottles and the total volume of Water , as well as how to solve for volume and the number of bottles.

Equation Relating Bottles and Volume

The relationship between the number of bottles and the total volume of water can be represented by the equation: Volume = Constant Proportionality × Number of Bottles. Here, volume is measured in fluid ounces, and the constant proportionality represents the volume per bottle.

For example, if each bottle holds 17 fluid ounces, the equation becomes: Volume = 17 fluid ounces × Number of Bottles.

Solving for Volume and Number of Bottles

To find the volume or the number of bottles based on a given value, we can use the equation: Volume or Number of Bottles = Constant Proportionality × Number of Bottles or Volume.

Suppose we want to determine the volume for 51 bottles. Using the equation, we find that the volume is 867 fluid ounces.

Conversely, if we have a volume of 51 fluid ounces and want to know the number of bottles required to hold that volume, we can rearrange the equation. By dividing 51 fluid ounces by 17 fluid ounces per bottle, we find that we need 3 bottles.

Conversion of Kilometers to Miles

In this section, we will discuss the conversion of kilometers to miles and the proportional relationship between the two. We will explore the equation relating distance measured in kilometers to distance measured in miles and how to solve for distance.

Proportional Relationship between Kilometers and Miles

One mile is approximately equal to 1.61 kilometers. This means there is a constant proportionality between the distance measured in kilometers and the distance measured in miles.

To represent this relationship, we can use the equation: Distance in Miles = Constant Proportionality × Distance in Kilometers.

Writing Equations and Solving for Distance

If we want to convert a given distance from kilometers to miles, we can use the equation Distance in Miles = 1.61 × Distance in Kilometers.

For example, if we want to convert a distance of 10 kilometers to miles, we can plug in the values into the equation: Distance in Miles = 1.61 × 10. By solving the equation, we find that the distance is approximately 16.1 miles.

In this section, we will explore the proportional relationship between quarters and toonies. A toonie is a Canadian coin worth 2 dollars, and we can determine the number of quarters equivalent to a specific number of toonies.

For example, to find the number of quarters equivalent to 16 toonies, we can use the equation: Quarters = 16 × (1/8) = 2.

In this section, we will complete tables and identify constants of proportionality. By filling in missing parts of the table and observing Patterns , we can determine the constant proportionality for each case.

For example, if we have a table with different values of x and y and observe that the ratio between y and x remains the same, we can identify the constant of proportionality. This constant represents the relationship between the variables and helps us Create equations and solve problems.

In this section, we will discuss the process of determining whether polygons are scaled copies or not. By examining the angles and side lengths of the polygons, we can identify similarities or differences that indicate if they are scaled copies.

For example, if two polygons have different angles or side lengths, they are not scaled copies. However, if the angles and side lengths are proportional, they may be scaled copies.

In this article, we covered various topics related to using equations to solve problems. We explored the concepts of organizing and ordering values, proportional relationships, and applying equations to solve real-life scenarios such as ticket sales and recycling. By understanding these concepts and mastering the use of equations, one can effectively solve a wide range of problems.

The above is a brief introduction to Master Math 7 with Morgan: Get Help with Homework!

Let's move on to the first section of Master Math 7 with Morgan: Get Help with Homework!

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