The length of a rectangular piece of carpet is 2 meters less than 5 times its width. Find the dimensions of the carpet if its area is 16 square meters. Solve algebraically.
w = width
5w - 2 = length {length is 2 less than 5 times width}
Area of a rectangle = length x width
w(5w - 2) = 16 {substituted length, width, and given area into area formula}
5w² - 2w = 16 {used distributive property}
5w² - 2w - 16 = 0 {subtracted 16 from each side}
5w² - 10w + 8w - 16 = 0 {split -2w into -10w and 8w, factoring by grouping}
5w(w - 2) + 8(w - 2) = 0 {factored 5w out of first two terms and 8 out of last two terms}
(5w + 8)(w - 2) = 0 {factored out the common factor of the two terms, (w - 2)}
5w + 8 = 0 or w - 2 = 0 {set each factor equal to 0}
w = -8/5 ow w = 2 {solved each equation}
w = 2 {width cannot be negative}
5w - 2 = 8 {substituted 2, in for w, into 5w - 2}
width = 2m
length = 8m
- Algebra House
5w - 2 = length {length is 2 less than 5 times width}
Area of a rectangle = length x width
w(5w - 2) = 16 {substituted length, width, and given area into area formula}
5w² - 2w = 16 {used distributive property}
5w² - 2w - 16 = 0 {subtracted 16 from each side}
5w² - 10w + 8w - 16 = 0 {split -2w into -10w and 8w, factoring by grouping}
5w(w - 2) + 8(w - 2) = 0 {factored 5w out of first two terms and 8 out of last two terms}
(5w + 8)(w - 2) = 0 {factored out the common factor of the two terms, (w - 2)}
5w + 8 = 0 or w - 2 = 0 {set each factor equal to 0}
w = -8/5 ow w = 2 {solved each equation}
w = 2 {width cannot be negative}
5w - 2 = 8 {substituted 2, in for w, into 5w - 2}
width = 2m
length = 8m
- Algebra House
