Question

When the width of a rectangle with a length of 3/5 foot was decreased by 1/3 foot, the area of the rectangle became 7/25. Find the original width of the rectangle.
Plz help ASAP

Answer 1

Answer:

$\large\boxed{\text{The original width}\ =\dfrac{4}{5}\ ft}$

Step-by-step explanation:

$\text{The dimensions of rectangle:}\ \dfrac{3}{5}\times w.\\\\\text{The dimensions of new rectangel:}\ \dfrac{3}{5}\times\left(w-\dfrac{1}{3}\right)\\\\\text{The area of the new rectangle:}\ A=\dfrac{7}{25}\ ft^2\\\\\text{We have the equation:}\\\\\dfrac{3}{5}\left(w-\dfrac{1}{3}\right)=\dfrac{7}{25}\qquad\text{multiply both sides by 25}\\\\25\!\!\!\!\!\diagup^5\cdot\dfrac{3}{5\!\!\!\!\diagup_1}\left(w-\dfrac{1}{3}\right)=25\!\!\!\!\!\diagup^1\cdot\dfrac{7}{25\!\!\!\!\!\diagup_1}$

$15\left(w-\dfrac{1}{3}\right)=7\qquad\text{use the distributive property}\\\\15w-15\!\!\!\!\!\diagup^5\cdot\dfrac{1}{3\!\!\!\!\diagup_1}=7\\\\15w-5=7\qquad\text{add 5 to both sides}\\\\15w=12\qquad\text{divide both sides by 15}\\\\w=\dfrac{12:3}{15:3}\\\\w=\dfrac{4}{5}$