Question 1129247
length: {{{L}}}
width:{{{W}}}

if the    length    of    a    rectangle    is    {{{7}}}    meters    less    than    {{{twice }}}   the    width, we have    

{{{L=2W-7}}}............eq.1

     
if the    area    is   {{{ 60}}}    square    meters, we have

{{{L*W=60}}}......eq.2 ....substitute {{{L}}}

{{{(2W-7)*W=60}}}

{{{2W^2-7W=60}}}

{{{2W^2-7W-60=0}}}

{{{2W^2-15W+8W-60=0}}}

{{{(2W^2+8W)-(15W+60)=0}}}

{{{2W(W+4)-15(W+4)=0}}}

{{{(W + 4) (2W - 15) = 0}}}

since width, we need only positive solution

{{{(2 W - 15) = 0}}}=>{{{W=15/2}}} =>{{{W=7.5}}}


{{{L=2W-7}}}............eq.1

{{{L=2*7.5-7}}}
{{{L=15-7}}}
{{{L=8}}}