Question 1170662

if the length {{{L}}} of a rectangular garden is two feet less than three times the width {{{W}}} , we have

{{{L=3W-2}}} ....eq.1

if the area is {{{65ft^2, we have

{{{L*W=65}}}....eq.2

substitute {{{L}}} from eq.1

{{{(3W-2)*W=65}}}....solve for {{{W}}}

{{{3W^2-2W=65}}}

{{{3W^2-2W-65=0}}}....factor

{{{3W^2+13W-15W-65=0}}}

{{{(3W^2-15W)+(13W-65)=0}}}

{{{3W(W-5)+13(W-5)=0}}}

{{{(W - 5) (3W + 13) = 0}}}

use only positive solution

->{{{(W - 5) = 0}}}=>{{{W=5}}}

go to eq.1 

{{{L=3W-2}}} ....eq.1, substitute {{{W}}}

{{{L=3*5-2}}} 

{{{L=13}}} 

  dimension:

the length of a rectangular garden is {{{13ft}}} 

the width of a rectangular garden is {{{5ft}}}