Step-by-step explanation
Given that the area of the rectangular certificate is 35 inches and its perimeter is 24 inches. Therefore, if L represents the length of the certificate and w represents its width, therefore;
![\begin{gathered} lw=35---1 \\ 2(l+w)=24---2 \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/vcu1ebfdf4ric767uel695bkrsphg8jjnz.png)
Therefore, we can say
![l=(35)/(w)](https://img.qammunity.org/2023/formulas/mathematics/college/ldzgr2eqy4qn3lts90yckxhgqgp78adbct.png)
We will substitute the above in equation 2
![\begin{gathered} 2((35)/(w)+w)=24 \\ (70)/(w)+2w=24 \\ multiply\text{ though by w} \\ 70+2w^2=24w \\ 2w^2-24w+70=0 \\ 2(w^2-12w+35)=0 \\ w^2-7w-5w+35=0 \\ (w-7)-5(w-7)=0 \\ (w-7)(w-5)=0 \\ w=7\text{ or w=5} \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/ekzmq8t9dlnjxommha74cm0qzbmdh09z5q.png)
Since the width must be shorter than the length therefore the width will be 5 inches.
Hence;
![l=(35)/(5)=7](https://img.qammunity.org/2023/formulas/mathematics/college/mywhl350h3gf181uondedgle3bm63oj2qr.png)
Answers:
The dimensions are:
Length = 7 inches
Width = 5 inches