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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 58

Find the length and width of a rectangle whose perimeter is 40 feet and whose area is 96 square feet.

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Let the length of the rectangle be \(L\) feet and the width be \(W\) feet. We are given two conditions: the perimeter and the area.
Write the equation for the perimeter of a rectangle: \$2L + 2W = 40\(. Simplify this to \)L + W = 20$.
Write the equation for the area of a rectangle: \(L \times W = 96\).
From the perimeter equation, express one variable in terms of the other, for example, \(L = 20 - W\).
Substitute \(L = 20 - W\) into the area equation to get \((20 - W) \times W = 96\). This will give a quadratic equation in terms of \(W\) that you can solve.

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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 as P = 2(length + width). This formula helps relate the length and width when the perimeter is known, providing one equation to solve for the dimensions.

Area of a Rectangle

The area of a rectangle is the amount of space inside it, found by multiplying length by width (A = length × width). Knowing the area gives a second equation that, combined with the perimeter, allows solving for both dimensions.
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Solving Systems of Equations

To find the length and width, you set up two equations from the perimeter and area formulas and solve them simultaneously. Techniques include substitution or elimination, which help find the values of length and width that satisfy both conditions.
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