Let the length be x and the width be y.
As per the first condition:
[tex]x=3y-2\ldots(i)[/tex]As per the second condition, the area is 65 sq feet so it follows:
[tex]xy=65\ldots(ii)[/tex]Substitute the value of x obtained from (i) in (ii) and solve for y as follows:
[tex]\begin{gathered} y(3y-2)=65 \\ 3y^2-2y=65 \\ 3y^2-2y-65=0 \end{gathered}[/tex]Solve the quadratic as follows:
[tex]\begin{gathered} 3y^2-15y+13y-65=0 \\ 3y(y-5)+13(y-5)=0 \\ (y-5)(3y+13)=0 \\ y=5,-\frac{13}{3} \end{gathered}[/tex]Since the width is never negative, y=5, therefore solve for x from (i) to get:
[tex]x=3(5)-2=13[/tex]Hence the length is 13 feet and the width is 5 feet.
