Answer:
[tex]\large\boxed{\text{The original width}\ =\dfrac{4}{5}\ ft}[/tex]
Step-by-step explanation:
[tex]\text{The dimensions of rectangle:}\ \dfrac{3}{5}\times w.\\\\\text{The dimensions of new rectangel:}\ \dfrac{3}{5}\times\left(w-\dfrac{1}{3}\right)\\\\\text{The area of the new rectangle:}\ A=\dfrac{7}{25}\ ft^2\\\\\text{We have the equation:}\\\\\dfrac{3}{5}\left(w-\dfrac{1}{3}\right)=\dfrac{7}{25}\qquad\text{multiply both sides by 25}\\\\25\!\!\!\!\!\diagup^5\cdot\dfrac{3}{5\!\!\!\!\diagup_1}\left(w-\dfrac{1}{3}\right)=25\!\!\!\!\!\diagup^1\cdot\dfrac{7}{25\!\!\!\!\!\diagup_1}[/tex]
[tex]15\left(w-\dfrac{1}{3}\right)=7\qquad\text{use the distributive property}\\\\15w-15\!\!\!\!\!\diagup^5\cdot\dfrac{1}{3\!\!\!\!\diagup_1}=7\\\\15w-5=7\qquad\text{add 5 to both sides}\\\\15w=12\qquad\text{divide both sides by 15}\\\\w=\dfrac{12:3}{15:3}\\\\w=\dfrac{4}{5}[/tex]
