Question

I need some help with this question cause i didnt quite understand it

Answer 1

Consider that the area (A) of a rectangle with length L and width W, is given by,

$A=L\times W$

According to the problem, it is given that,

$\begin{gathered} A=x^2-x-72 \\ L=x+8 \end{gathered}$

Substitute the values and solve for W as,

$\begin{gathered} x^2-x-72=(x+8)\times W \\ W=\frac{x^2-x-72}{x+8} \\ W=\frac{x^2-9x+8x-72}{x+8} \\ W=\frac{x^{}(x-9)+8(x-9)}{x+8} \\ W=\frac{^{}(x-9)(x+8)}{x+8} \\ W=x-9 \end{gathered}$

Thus, the width of the rectangle is (x-9) meters.