Table of contents
- 0. Review of Algebra4h 16m
- 1. Equations & Inequalities3h 18m
- 2. Graphs of Equations43m
- 3. Functions2h 17m
- 4. Polynomial Functions1h 44m
- 5. Rational Functions1h 23m
- 6. Exponential & Logarithmic Functions2h 28m
- 7. Systems of Equations & Matrices4h 6m
- 8. Conic Sections2h 23m
- 9. Sequences, Series, & Induction1h 19m
- 10. Combinatorics & Probability1h 45m
1. Equations & Inequalities
Linear Equations
3:32 minutes
Problem 43a
Textbook Question
Textbook QuestionIn Exercises 36–43, use the five-step strategy for solving word problems. The length of a rectangular field is 6 yards less than triple the width. If the perimeter of the field is 340 yards, what are its dimensions?
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Key Concepts
Here are the essential concepts you must grasp in order to answer the question correctly.
Perimeter of a Rectangle
The perimeter of a rectangle is the total distance around the rectangle, calculated by the formula P = 2L + 2W, where L is the length and W is the width. Understanding this formula is essential for solving problems related to the dimensions of rectangular shapes, as it allows you to relate the length and width to the given perimeter.
Algebraic Expressions
Algebraic expressions are mathematical phrases that can include numbers, variables, and operations. In this problem, the relationship between the length and width of the field is expressed algebraically, where the length is defined as 6 yards less than triple the width. This understanding is crucial for setting up equations that represent the problem.
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Solving Linear Equations
Solving linear equations involves finding the value of variables that satisfy the equation. In this context, once the relationships between length and width are established, you will set up and solve a linear equation based on the perimeter. Mastery of this concept is vital for determining the dimensions of the rectangular field.
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