Question 1151179

if the length of a rectangle is {{{1}}} yd less than {{{twice}}} the {{{width}}}, we have

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

 and, if the area of the rectangle is {{{21}}} yd^2, we have

{{{L*W=21}}}.......solve for {{{L}}}

{{{L=21/W}}}....eq.2

from eq.1 and eq.2 we have

{{{2W-1=21/W}}}......solve for {{{W}}}

{{{2W*W-W=21}}}

{{{2W^2-W-21=0}}}.........factor

{{{2W^2+6W-7W-21=0}}}

{{{(2W^2+6W)-(7W+21)=0}}}

{{{2W(W+3)-7(W+3)=0}}}

{{{(W + 3) (2W - 7) = 0}}}


solutions:

if {{{(W + 3)  = 0}}}=> {{{W=-3}}}=> since width, disregard negative solution

if {{{ (2W - 7) = 0}}}=> {{{W=3.5}}} yd

{{{L=6}}} yd


so, the dimensions of the rectangle are:

the length={{{6}}} yd
the width={{{3.5}}} yd