Question 1046123

The length {{{L}}} of a rectangle is {{{2in}}} greater than its width {{{W}}}.

{{{L=W+2in}}}
The perimeter {{{P=2(L+W)}}} is {{{36in}}}, so {{{2(L+W)=36in}}}.

{{{2(L+W)=36in}}}....substitute {{{L}}}

{{{2(W+2in+W)=36in}}}.....solve for {{{W}}}

{{{(W+2in+W)=36in/2}}}

{{{2W+2in=18in}}}

{{{2W=18in-2in}}}

{{{2W=16in}}}

{{{highlight(W=8in)}}}

now find {{{L}}}

{{{L=W+2in}}}

{{{L=8in+2in}}}

{{{highlight(L=10in)}}}

so, the dimensions of the rectangle are: 

the length is {{{highlight(10in)}}} 
and 
the width is {{{highlight(8in)}}}