Question 993198

let the length be {{{L}}} and the width {{{W}}}

if the length of a rectangle is {{{3in}}} more than its width, than we have

{{{L=W+3in}}}....eq.1


if the perimeter of the rectangle is {{{P=30in }}}, we have

{{{2(L+W)=30in }}}....substitute {{{L}}} from eq.1

{{{2(W+3in+W)=30in }}}.......solve for {{{W}}}

{{{(2W+3in)=30in/2 }}}

{{{2W+3in=15in }}}

{{{2W=15in-3in }}}

{{{2W=12in }}}

{{{W=12in/2 }}}

{{{highlight(W=6in) }}}

go back to {{{L=W+3in}}}....eq.1 substitute {{{6in}}} for substitute {{{W}}} and solve for substitute {{{L}}}

{{{L=6in+3in}}}

{{{highlight(L=9in)}}}