Let the width of the rectangle be 'w'
According to the question length of the rectangle (l) is three feet more than twice it's width;
[tex] \longrightarrow [/tex] l = (2w + 3) ft
Perimeter of rectangle = 78 ft
Formula of perimeter of rectangle = 2(Length + Width)
So,
[tex] \rm \implies 2(Length + Width) = 78 \\ \\ \rm \implies 2(2w + 3 + w) = 78 \\ \\ \rm \implies 2(3w + 3) = 78 \\ \\ \rm \implies 2 \times 3(w + 1) = 78 \\ \\ \rm \implies 6(w + 1) = 78 \\ \\ \rm \implies w + 1 = \dfrac{78}{6} \\ \\ \rm \implies w + 1 = 13 \\ \\ \rm \implies w = 13 - 1 \\ \\ \rm \implies w = 12 \: ft[/tex]
[tex] \therefore [/tex] Width of the rectangle (w) = 12 ft
