Learn on PengiSaxon Math, Course 3Chapter 7: Algebra

Lesson 70: Solving Direct Variation Problems

In this Grade 8 Saxon Math Course 3 lesson, students learn to solve direct variation problems by identifying the constant of proportionality (k) and using it to write equations and proportions relating two quantities that vary directly. Students practice finding k from a given pair of values, then applying it to calculate unknown values in real-world contexts such as pricing, distance-rate-time, and wait-time scenarios.

Section 1

📘 Solving Direct Variation Problems

New Concept

When two quantities vary directly, they are proportional. We use the letter kk for the constant of proportionality, and their relationship is defined as:

quantity Aquantity B=k\frac{\text{quantity A}}{\text{quantity B}} = k

What’s next

This card is just the foundation. Soon, you'll master solving for unknown values using both the constant of proportionality and setting up direct proportions.

Section 2

Understanding Direct Variation

Property

When two quantities vary directly, their ratio is a constant, which we call the constant of proportionality, kk.

Examples

If 2 CDs cost 30 dollars, the constant is k=302=15k = \frac{30}{2} = 15. So, 4 CDs will cost 4×15=604 \times 15 = 60 dollars.
Ellen charges 780 dollars for 60 people, so her rate is k=78060=13k = \frac{780}{60} = 13 dollars per person. For 100 people, she charges 100×13=1300100 \times 13 = 1300 dollars.
A spring stretches 2 cm for a 30g weight, so k=230=115k = \frac{2}{30} = \frac{1}{15} cm/g. For a 75g weight, it stretches 75×115=575 \times \frac{1}{15} = 5 cm.

Explanation

Think of this as a 'buy one, get one for a fixed price' deal, but for any amount! The relationship between two things stays perfectly consistent. Once you find the special constant, kk, you can predict the outcome for any number. It's like having a secret formula to know the price of 100 CDs without counting!

Section 3

Solving With Proportions

Property

An alternate method for solving direct variation problems is to write a proportion by setting the two ratios equal to each other.

a1b1=a2b2 \frac{a_1}{b_1} = \frac{a_2}{b_2}

Examples

If 60 people cost 780 dollars, how much for 100? Set up the proportion: 78060=x100\frac{780}{60} = \frac{x}{100}. Cross-multiply to find x=1300x = 1300 dollars.
If Maggie drives 75 miles in 1.5 hours, how long for 100 miles? 751.5=100t\frac{75}{1.5} = \frac{100}{t}. Cross-multiply to find t=2t = 2 hours.
If Tom types 100 words in 2.5 minutes, how many in 15? 1002.5=w15\frac{100}{2.5} = \frac{w}{15}. Cross-multiply to find w=600w = 600 words.

Explanation

This is the ultimate shortcut for direct variation problems. You don't even need to find the constant kk! Just set up two fractions that you know are equal: one with the information you have, and one with the missing piece you want to find. Then, use the magic of cross-multiplication to solve for the unknown value.

Section 4

Predicting With Rates

Property

To make a prediction, first calculate the average rate of change. Then, multiply this rate by the remaining quantity to estimate the outcome.

Rate=Change in Quantity AChange in Quantity B \text{Rate} = \frac{\text{Change in Quantity A}}{\text{Change in Quantity B}}

Examples

In 3 minutes, 5 people were helped. The rate is 35=0.6\frac{3}{5} = 0.6 minutes per person. For the next 18 people, the wait is 18×0.6=10.818 \times 0.6 = 10.8 minutes.
Mitchell traveled 3 miles in 15 minutes. His rate is 153=5\frac{15}{3} = 5 minutes per mile. To travel 5 more miles, it will take him 5×5=255 \times 5 = 25 minutes.

Explanation

Life isn't always perfectly predictable, but you can make a solid estimate! First, figure out the 'speed' at which something is happening, like minutes per person in a line. Then, multiply that rate by how much is left to do. It’s like being a detective, using clues from the immediate past to solve a near-future mystery.

Book overview

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

Continue this chapter

Chapter 7: Algebra

  1. Lesson 1

    Lesson 61: Sequences

  2. Lesson 2

    Lesson 62: Graphing Solutions to Inequalities on a Number Line

  3. Lesson 3

    Lesson 63: Rational Numbers, Non-Terminating Decimals, and Percents and Fractions with Negative Exponents

  4. Lesson 4

    Lesson 64: Using a Unit Multiplier to Convert a Rate

  5. Lesson 5

    Lesson 65: Applications Using Similar Triangles

  6. Lesson 6

    Lesson 66: Special Right Triangles

  7. Lesson 7

    Lesson 67: Percent of Change

  8. Lesson 8

    Lesson 68: Probability Multiplication Rule

  9. Lesson 9

    Lesson 69: Direct Variation

  10. Lesson 10Current

    Lesson 70: Solving Direct Variation Problems

  11. Lesson 11

    Investigation 7: Probability Simulation

Lesson overview

Expand to review the lesson summary and core properties.

Expand

Section 1

📘 Solving Direct Variation Problems

New Concept

When two quantities vary directly, they are proportional. We use the letter kk for the constant of proportionality, and their relationship is defined as:

quantity Aquantity B=k\frac{\text{quantity A}}{\text{quantity B}} = k

What’s next

This card is just the foundation. Soon, you'll master solving for unknown values using both the constant of proportionality and setting up direct proportions.

Section 2

Understanding Direct Variation

Property

When two quantities vary directly, their ratio is a constant, which we call the constant of proportionality, kk.

Examples

If 2 CDs cost 30 dollars, the constant is k=302=15k = \frac{30}{2} = 15. So, 4 CDs will cost 4×15=604 \times 15 = 60 dollars.
Ellen charges 780 dollars for 60 people, so her rate is k=78060=13k = \frac{780}{60} = 13 dollars per person. For 100 people, she charges 100×13=1300100 \times 13 = 1300 dollars.
A spring stretches 2 cm for a 30g weight, so k=230=115k = \frac{2}{30} = \frac{1}{15} cm/g. For a 75g weight, it stretches 75×115=575 \times \frac{1}{15} = 5 cm.

Explanation

Think of this as a 'buy one, get one for a fixed price' deal, but for any amount! The relationship between two things stays perfectly consistent. Once you find the special constant, kk, you can predict the outcome for any number. It's like having a secret formula to know the price of 100 CDs without counting!

Section 3

Solving With Proportions

Property

An alternate method for solving direct variation problems is to write a proportion by setting the two ratios equal to each other.

a1b1=a2b2 \frac{a_1}{b_1} = \frac{a_2}{b_2}

Examples

If 60 people cost 780 dollars, how much for 100? Set up the proportion: 78060=x100\frac{780}{60} = \frac{x}{100}. Cross-multiply to find x=1300x = 1300 dollars.
If Maggie drives 75 miles in 1.5 hours, how long for 100 miles? 751.5=100t\frac{75}{1.5} = \frac{100}{t}. Cross-multiply to find t=2t = 2 hours.
If Tom types 100 words in 2.5 minutes, how many in 15? 1002.5=w15\frac{100}{2.5} = \frac{w}{15}. Cross-multiply to find w=600w = 600 words.

Explanation

This is the ultimate shortcut for direct variation problems. You don't even need to find the constant kk! Just set up two fractions that you know are equal: one with the information you have, and one with the missing piece you want to find. Then, use the magic of cross-multiplication to solve for the unknown value.

Section 4

Predicting With Rates

Property

To make a prediction, first calculate the average rate of change. Then, multiply this rate by the remaining quantity to estimate the outcome.

Rate=Change in Quantity AChange in Quantity B \text{Rate} = \frac{\text{Change in Quantity A}}{\text{Change in Quantity B}}

Examples

In 3 minutes, 5 people were helped. The rate is 35=0.6\frac{3}{5} = 0.6 minutes per person. For the next 18 people, the wait is 18×0.6=10.818 \times 0.6 = 10.8 minutes.
Mitchell traveled 3 miles in 15 minutes. His rate is 153=5\frac{15}{3} = 5 minutes per mile. To travel 5 more miles, it will take him 5×5=255 \times 5 = 25 minutes.

Explanation

Life isn't always perfectly predictable, but you can make a solid estimate! First, figure out the 'speed' at which something is happening, like minutes per person in a line. Then, multiply that rate by how much is left to do. It’s like being a detective, using clues from the immediate past to solve a near-future mystery.

Book overview

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

Continue this chapter

Chapter 7: Algebra

  1. Lesson 1

    Lesson 61: Sequences

  2. Lesson 2

    Lesson 62: Graphing Solutions to Inequalities on a Number Line

  3. Lesson 3

    Lesson 63: Rational Numbers, Non-Terminating Decimals, and Percents and Fractions with Negative Exponents

  4. Lesson 4

    Lesson 64: Using a Unit Multiplier to Convert a Rate

  5. Lesson 5

    Lesson 65: Applications Using Similar Triangles

  6. Lesson 6

    Lesson 66: Special Right Triangles

  7. Lesson 7

    Lesson 67: Percent of Change

  8. Lesson 8

    Lesson 68: Probability Multiplication Rule

  9. Lesson 9

    Lesson 69: Direct Variation

  10. Lesson 10Current

    Lesson 70: Solving Direct Variation Problems

  11. Lesson 11

    Investigation 7: Probability Simulation