Learn on PengiSaxon Algebra 2Chapter 1: Lessons 1-10, Investigation 1

Lesson 8: Finding Direct Variation

In Saxon Algebra 2 Lesson 8, Grade 10 students learn to identify and solve direct variation relationships using the equation A = kB, where k is the constant of variation. The lesson covers two solution methods: finding the constant of variation to build an equation, and setting up equivalent ratios as a proportion. Students apply these skills to real-world contexts including distance-time relationships, unit conversions, and gas volume and temperature using Charles's Law.

Section 1

๐Ÿ“˜ Finding Direct Variation

New Concept

When a problem states that AA varies directly as BB, the equation A=kBA = kB is implied. This relationship is called direct variation.

Why it matters

Direct variation is your first tool for translating real-world patterns into the language of algebra. Mastering this simple linear relationship is the foundation for modeling everything from chemical reactions to economic trends.

Whatโ€™s next

Next, youโ€™ll translate word problems into the equation A=kBA=kB, find the constant of variation, and solve for unknown values.

Section 2

direct variation

When the statement of a problem says that AA varies directly as BB or that AA is directly proportional to BB, the equation A=kBA = kB is implied.

  1. If pay (PP) varies directly with hours worked (HH), the relationship is P=kHP = kH. If you earn 50 dollars for 5 hours, you'll earn 100 dollars for 10 hours. 2. The resistance (RR) is directly proportional to the length (LL), so R=kLR=kL. A longer wire means proportionally higher resistance.

Think of this as a perfect partnership! When one value changes, the other changes by the exact same multiplier. If you buy more pizza slices, the cost goes up predictably. Itโ€™s a straight-line relationship!

Section 3

constant of variation

The constant kk in the equation A=kBA = kB is called the constant of variation.

  1. A scooter travels 10 km in 30 minutes (D=kTD=kT). To find k, set up 10=k(30)10 = k(30), so the constant is k=13k = \frac{1}{3}. 2. If 120 seconds have passed in 2 minutes (S=kMS=kM), then 120=k(2)120 = k(2), so the constant of variation is k=60k = 60.

Meet 'k', the secret code of the relationship! This number is the specific multiplier that never changes for a given situation. Find 'k' first, and you can solve any problem related to that variation.

Section 4

Direct variation as a ratio

The statement AA varies directly as BB also implies the proportion:

A1A2=B1B2\frac{A_1}{A_2} = \frac{B_1}{B_2}

  1. If 12 items cost 78 dollars, how much for 42 items? 78C2=1242โ†’78โ‹…42=12โ‹…C2โ†’C2=273\frac{78}{C_2} = \frac{12}{42} \rightarrow 78 \cdot 42 = 12 \cdot C_2 \rightarrow C_2 = 273 dollars. 2. If 15 ties are for 30 feet of track, how many for 50 feet? 15T2=3050โ†’15โ‹…50=30โ‹…T2โ†’T2=25\frac{15}{T_2} = \frac{30}{50} \rightarrow 15 \cdot 50 = 30 \cdot T_2 \rightarrow T_2 = 25 ties.

This is a fantastic shortcut for solving problems when you don't need to find the constant 'k'. Just set up the two pairs of values as a proportion, cross-multiply, and you've found your answer!

Book overview

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

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Chapter 1: Lessons 1-10, Investigation 1

  1. Lesson 1

    Lesson 1: Using Properties of Real Numbers

  2. Lesson 2

    Lesson 2: Evaluating Expressions and Combining Like Terms

  3. Lesson 3

    Lesson 3: Using Rules of Exponents

  4. Lesson 4

    LAB 1: Graphing Calculator: Graphing a Function and Building a Table

  5. Lesson 5

    Lesson 4: Identifying Functions and Using Function Notation

  6. Lesson 6

    LAB 2: Graphing Calculator: Storing and Recalling Data in a Matrix

  7. Lesson 7

    Lesson 5: Using Matrices to Organize Data and to Solve Problems

  8. Lesson 8

    Lesson 6: Finding Percent of Change

  9. Lesson 9

    Lesson 7: Solving Linear Equations (Exploration: Solving Equations Using Algebra Tiles)

  10. Lesson 10Current

    Lesson 8: Finding Direct Variation

  11. Lesson 11

    Lesson 9: Multiplying Matrices

  12. Lesson 12

    Lesson 10: Solving and Graphing Inequalities

  13. Lesson 13

    Investigation 1: Logic and Truth Tables

Lesson overview

Expand to review the lesson summary and core properties.

Expand

Section 1

๐Ÿ“˜ Finding Direct Variation

New Concept

When a problem states that AA varies directly as BB, the equation A=kBA = kB is implied. This relationship is called direct variation.

Why it matters

Direct variation is your first tool for translating real-world patterns into the language of algebra. Mastering this simple linear relationship is the foundation for modeling everything from chemical reactions to economic trends.

Whatโ€™s next

Next, youโ€™ll translate word problems into the equation A=kBA=kB, find the constant of variation, and solve for unknown values.

Section 2

direct variation

When the statement of a problem says that AA varies directly as BB or that AA is directly proportional to BB, the equation A=kBA = kB is implied.

  1. If pay (PP) varies directly with hours worked (HH), the relationship is P=kHP = kH. If you earn 50 dollars for 5 hours, you'll earn 100 dollars for 10 hours. 2. The resistance (RR) is directly proportional to the length (LL), so R=kLR=kL. A longer wire means proportionally higher resistance.

Think of this as a perfect partnership! When one value changes, the other changes by the exact same multiplier. If you buy more pizza slices, the cost goes up predictably. Itโ€™s a straight-line relationship!

Section 3

constant of variation

The constant kk in the equation A=kBA = kB is called the constant of variation.

  1. A scooter travels 10 km in 30 minutes (D=kTD=kT). To find k, set up 10=k(30)10 = k(30), so the constant is k=13k = \frac{1}{3}. 2. If 120 seconds have passed in 2 minutes (S=kMS=kM), then 120=k(2)120 = k(2), so the constant of variation is k=60k = 60.

Meet 'k', the secret code of the relationship! This number is the specific multiplier that never changes for a given situation. Find 'k' first, and you can solve any problem related to that variation.

Section 4

Direct variation as a ratio

The statement AA varies directly as BB also implies the proportion:

A1A2=B1B2\frac{A_1}{A_2} = \frac{B_1}{B_2}

  1. If 12 items cost 78 dollars, how much for 42 items? 78C2=1242โ†’78โ‹…42=12โ‹…C2โ†’C2=273\frac{78}{C_2} = \frac{12}{42} \rightarrow 78 \cdot 42 = 12 \cdot C_2 \rightarrow C_2 = 273 dollars. 2. If 15 ties are for 30 feet of track, how many for 50 feet? 15T2=3050โ†’15โ‹…50=30โ‹…T2โ†’T2=25\frac{15}{T_2} = \frac{30}{50} \rightarrow 15 \cdot 50 = 30 \cdot T_2 \rightarrow T_2 = 25 ties.

This is a fantastic shortcut for solving problems when you don't need to find the constant 'k'. Just set up the two pairs of values as a proportion, cross-multiply, and you've found your answer!

Book overview

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

Continue this chapter

Chapter 1: Lessons 1-10, Investigation 1

  1. Lesson 1

    Lesson 1: Using Properties of Real Numbers

  2. Lesson 2

    Lesson 2: Evaluating Expressions and Combining Like Terms

  3. Lesson 3

    Lesson 3: Using Rules of Exponents

  4. Lesson 4

    LAB 1: Graphing Calculator: Graphing a Function and Building a Table

  5. Lesson 5

    Lesson 4: Identifying Functions and Using Function Notation

  6. Lesson 6

    LAB 2: Graphing Calculator: Storing and Recalling Data in a Matrix

  7. Lesson 7

    Lesson 5: Using Matrices to Organize Data and to Solve Problems

  8. Lesson 8

    Lesson 6: Finding Percent of Change

  9. Lesson 9

    Lesson 7: Solving Linear Equations (Exploration: Solving Equations Using Algebra Tiles)

  10. Lesson 10Current

    Lesson 8: Finding Direct Variation

  11. Lesson 11

    Lesson 9: Multiplying Matrices

  12. Lesson 12

    Lesson 10: Solving and Graphing Inequalities

  13. Lesson 13

    Investigation 1: Logic and Truth Tables