Learn on PengiSaxon Math, Course 3Chapter 5: Number & Operations • Algebra

Lesson 49: Solving Rate Problems with Proportions and Equations

In this Grade 8 Saxon Math Course 3 lesson, students learn to solve rate problems using two methods: setting up proportions with ratio tables and multiplying by the unit rate. The lesson introduces the two core rate equations — y/x = k and y = kx — and applies them to real-world scenarios involving weight, speed, and ticket pricing. Students practice identifying unit rates, writing proportions, and using cross multiplication to find unknown quantities in proportional relationships.

Section 1

📘 Solving Rate Problems with Proportions and Equations

New Concept

Proportional relationships can be expressed with two key equations, allowing you to solve rate problems using either proportions or a direct formula.

All proportional relationships can be expressed with these two equations:

  1. The ratio is constant: yx=k \frac{y}{x} = k
  2. The unit rate is constant: y=kx y = kx

What’s next

You're now ready to see these methods in action. Next, you'll walk through worked examples applying both proportions and rate equations to solve the same problems.

Section 2

Unit Rate

Property

A unit rate is a rate with a denominator of 1, showing the value 'per one' item. The formula is k=yxk = \frac{y}{x}.

Examples

20.25 dollars3 tickets=6.75\frac{20.25 \text{ dollars}}{3 \text{ tickets}} = 6.75 dollars per ticket.
15 pounds6 books=2.5\frac{15 \text{ pounds}}{6 \text{ books}} = 2.5 pounds per book.

Explanation

Think of it as the price tag for just one! Knowing the unit rate helps you find the best deal or compare speeds. It’s your secret weapon for becoming a super-shopper and making smart choices!

Section 3

Solving with Proportions

Property

Proportions show that two ratios are equal. Use this when a relationship is proportional to find a missing value: ab=cd\frac{a}{b} = \frac{c}{d}.

Examples

6 books15 lbs=20 booksp lbs    p=50\frac{6 \text{ books}}{15 \text{ lbs}} = \frac{20 \text{ books}}{p \text{ lbs}} \implies p=50 lbs.
15 km27 min=50 kmm min    m=90\frac{15 \text{ km}}{27 \text{ min}} = \frac{50 \text{ km}}{m \text{ min}} \implies m=90 min.

Explanation

Think of this as your ultimate scaling tool! If you know the ratio for one situation, you can scale it to find a missing piece in another, keeping everything fair and balanced like a math seesaw.

Section 4

Solving with a Unit Rate Equation

Property

Use the equation y=kxy = kx, where kk is the constant unit rate. Multiply the rate (kk) by the quantity (xx) to find the total (yy).

Examples

Unit rate is 2.52.5 lbs/book. Weight of 20 books is w=2.520=50w = 2.5 \cdot 20 = 50 lbs.

Explanation

First, find the 'per one' value, then multiply away! This direct method is super fast once you have the unit rate. No complex fractions, just simple multiplication to get your answer.

Book overview

Jump across lessons in the current chapter without opening the full course modal.

Continue this chapter

Chapter 5: Number & Operations • Algebra

  1. Lesson 1

    Lesson 41: Functions

  2. Lesson 2

    Lesson 42: Volume

  3. Lesson 3

    Lesson 43: Surface Area

  4. Lesson 4

    Lesson 44: Solving Proportions Using Cross Products and Slope of a Line

  5. Lesson 5

    Lesson 45: Ratio Problems Involving Totals

  6. Lesson 6

    Lesson 46: Solving Problems Using Scientific Notation

  7. Lesson 7

    Lesson 47: Graphing Functions

  8. Lesson 8

    Lesson 48: Percent of a Whole

  9. Lesson 9Current

    Lesson 49: Solving Rate Problems with Proportions and Equations

  10. Lesson 10

    Lesson 50: Solving Multi-Step Equations

  11. Lesson 11

    Lesson 11: Graphing Transformations

Lesson overview

Expand to review the lesson summary and core properties.

Expand

Section 1

📘 Solving Rate Problems with Proportions and Equations

New Concept

Proportional relationships can be expressed with two key equations, allowing you to solve rate problems using either proportions or a direct formula.

All proportional relationships can be expressed with these two equations:

  1. The ratio is constant: yx=k \frac{y}{x} = k
  2. The unit rate is constant: y=kx y = kx

What’s next

You're now ready to see these methods in action. Next, you'll walk through worked examples applying both proportions and rate equations to solve the same problems.

Section 2

Unit Rate

Property

A unit rate is a rate with a denominator of 1, showing the value 'per one' item. The formula is k=yxk = \frac{y}{x}.

Examples

20.25 dollars3 tickets=6.75\frac{20.25 \text{ dollars}}{3 \text{ tickets}} = 6.75 dollars per ticket.
15 pounds6 books=2.5\frac{15 \text{ pounds}}{6 \text{ books}} = 2.5 pounds per book.

Explanation

Think of it as the price tag for just one! Knowing the unit rate helps you find the best deal or compare speeds. It’s your secret weapon for becoming a super-shopper and making smart choices!

Section 3

Solving with Proportions

Property

Proportions show that two ratios are equal. Use this when a relationship is proportional to find a missing value: ab=cd\frac{a}{b} = \frac{c}{d}.

Examples

6 books15 lbs=20 booksp lbs    p=50\frac{6 \text{ books}}{15 \text{ lbs}} = \frac{20 \text{ books}}{p \text{ lbs}} \implies p=50 lbs.
15 km27 min=50 kmm min    m=90\frac{15 \text{ km}}{27 \text{ min}} = \frac{50 \text{ km}}{m \text{ min}} \implies m=90 min.

Explanation

Think of this as your ultimate scaling tool! If you know the ratio for one situation, you can scale it to find a missing piece in another, keeping everything fair and balanced like a math seesaw.

Section 4

Solving with a Unit Rate Equation

Property

Use the equation y=kxy = kx, where kk is the constant unit rate. Multiply the rate (kk) by the quantity (xx) to find the total (yy).

Examples

Unit rate is 2.52.5 lbs/book. Weight of 20 books is w=2.520=50w = 2.5 \cdot 20 = 50 lbs.

Explanation

First, find the 'per one' value, then multiply away! This direct method is super fast once you have the unit rate. No complex fractions, just simple multiplication to get your answer.

Book overview

Jump across lessons in the current chapter without opening the full course modal.

Continue this chapter

Chapter 5: Number & Operations • Algebra

  1. Lesson 1

    Lesson 41: Functions

  2. Lesson 2

    Lesson 42: Volume

  3. Lesson 3

    Lesson 43: Surface Area

  4. Lesson 4

    Lesson 44: Solving Proportions Using Cross Products and Slope of a Line

  5. Lesson 5

    Lesson 45: Ratio Problems Involving Totals

  6. Lesson 6

    Lesson 46: Solving Problems Using Scientific Notation

  7. Lesson 7

    Lesson 47: Graphing Functions

  8. Lesson 8

    Lesson 48: Percent of a Whole

  9. Lesson 9Current

    Lesson 49: Solving Rate Problems with Proportions and Equations

  10. Lesson 10

    Lesson 50: Solving Multi-Step Equations

  11. Lesson 11

    Lesson 11: Graphing Transformations