Question 1131620
let the length be {{{L}}} and te width {{{W}}}

if the length of a rectangle is {{{7cm}}} {{{less}}} than {{{three}}}{{{ times}}} its {{{width}}}, we have

{{{L=3W-7}}} ........eq.1


 if its area is {{{20cm^2}}}, we have

 {{{L*W=20}}}........substitute {{{L}}} from eq.1

{{{L*W=20}}}

{{{(3W-7)*W=20}}}

{{{3W^2-7W=20}}}

{{{3W^2-7W-20=0}}}........factor

{{{3W^2-12W+5W-20=0}}}

{{{(3W^2-12W)+(5W-20)=0}}}

{{{3W(W-4)+5(W-4)=0}}}

{{{(W - 4) (3 W + 5) = 0}}}


solutions:

if {{{(W - 4)  = 0}}}=>{{{W=4}}}

if {{{ (3 W + 5) = 0}}}=>{{{3W=-5}}}=>{{{W=-5/3}}}-> disregard negative solution for the width


now find the length

{{{L=3W-7}}} ........eq.1
{{{L=3*4-7}}} 
{{{L=5}}} 

so, {{{L=5cm}}}  and {{{W=4cm}}}