Perimeter of a rectangle

Find the width of a rectangle whose width is one more than twice its length and whose perimeter is 26.
  
x = length
2x + 1 = width {width is one more than twice the length} 
  
Perimeter of a rectangle = 2(width) + 2(length)

2(x) + 2(2x + 1) = 26 {perimeter is 2(width) + 2(length)}
2x + 4x + 2 = 26 {used distributive property}
6x + 2 = 26 {combined like terms}
6x = 24 {subtracted 2 from both sides}
x = 4 {divided both sides by 4}
2x + 1 = 9 {substituted 4, in for x, into 2x + 1}
width = 4 and length = 9

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Find the width and length.

The width of a rectangle is 1 ft less than the length.
The area is 2 ft². Find the length and the width. 

x = length
x - 1 = width {width is one less than the length} 

Area of a rectangle is length x width. 

x(x - 1) = 2 {area is length times width}
x² - x = 2 {used distributive property}
x² - x - 2 = 0 {subtracted 2 from both sides}
(x - 2)(x + 1) = 0 {factored into two binomials}
x - 2 = 0 or x + 1 = 0 {set each factor equal to 0}
x = 2 or x = -1 {solved each equation for x}
x = 2 {length cannot be negative}
x - 1 = 1 {substituted 2, in for x, into x - 1} 

width = 1 ft and length = 2 ft

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