Question 1205994

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

perimeter is

{{{P=2(L+W)}}}

if a rectangle has a perimeter {{{90 }}}units, we have

{{{2(L+W)=90}}}

{{{L+W=45}}}.......eq.1


if the width is {{{6}}} more than {{{2}}} times the length, we have

{{{W=2L+6}}}....eq.2

substitute width in eq.1

{{{L+2L+6=45}}}.......solve for{{{ L}}}

{{{3L=45-6}}}

{{{3L=39}}}

{{{L=13}}}
 
 the length is {{{13}}} units