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Ch. 5 - Systems of Equations and Inequalities
Blitzer - College Algebra 8th Edition
Blitzer8th EditionCollege AlgebraISBN: 9780136970514Not the one you use?Change textbook
Chapter 6, Problem 57

Find the length and width of a rectangle whose perimeter is 36 feet and whose area is 77 square feet.

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Let the length of the rectangle be \(L\) feet and the width be \(W\) feet.
Write the equation for the perimeter of the rectangle: \$2L + 2W = 36$.
Simplify the perimeter equation by dividing both sides by 2: \(L + W = 18\).
Write the equation for the area of the rectangle: \(L \times W = 77\).
Use the perimeter equation to express \(L\) in terms of \(W\): \(L = 18 - W\), then substitute this into the area equation to get \((18 - W) \times W = 77\).

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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 it, calculated by adding twice the length and twice the width (P = 2L + 2W). Understanding this formula allows you to set up an equation relating length and width when the perimeter is known.

Area of a Rectangle

The area of a rectangle is the amount of space inside it, found by multiplying the length by the width (A = L × W). This formula helps create a second equation when the area is given, which is essential for solving for both dimensions.
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Solving Systems of Equations

When two equations involve two variables, such as length and width, solving the system means finding values that satisfy both simultaneously. Techniques include substitution or elimination, which are used here to find the rectangle’s dimensions from the perimeter and area equations.
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