Answer :
Explanation
We are given the following:
[tex]\begin{gathered} Let\text{ }the\text{ }shorter\text{ }side\text{ }be\text{ }S \\ then,\text{ }longer\text{ }side(L)=2S-3 \end{gathered}[/tex]We know that the area of a rectangle is calculated with the formula:
[tex]\begin{gathered} Area=L\times B \\ \therefore Area=L\times S \\ where\text{ }Area=170\text{ }inches^2 \end{gathered}[/tex]Therefore, we have:
[tex]\begin{gathered} \begin{equation*} Area=L\times S \end{equation*} \\ 170=(2S-3)\times S \\ 170=2S^2-3S \\ 2S^2-3S-170=0 \\ (2S+17)(S-10)=0 \\ 2S+17=0;S-10=0 \\ 2S=-17;S=10 \\ S=-\frac{17}{2};S=10 \\ Since\text{ }lengths\text{ }cannot\text{ }be\text{ }negative, \\ \therefore S=10inches \end{gathered}[/tex]The perimeter of the rectangle becomes:
[tex]\begin{gathered} Perimeter=2(L+B) \\ Perimeter=2(L+S) \\ where \\ L=2S-3=2(10)-3=17 \\ Perimeter=2(10+17) \\ Perimeter=2(27)=54inches \end{gathered}[/tex]Since, the solution must be algebraic, then we have:
[tex]\begin{gathered} Per\imaginaryI meter=2(L+S) \\ Per\imaginaryI meter=2(2S-3+S) \\ Per\imaginaryI meter=2(3S-3) \\ Per\imaginaryI meter=2\times3(S-1) \\ Per\imaginaryI meter=6(S-1) \end{gathered}[/tex]