Question 1126306

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

 A rectangle has an area {{{A=L*W}}}....

given:

{{{A=369in^2}}}=>{{{369in^2=L*W}}}

a length which is five inches more than four times the width, so we have

{{{L=4W+5in}}}....eq1


{{{369=L*W}}}...substitute {{{L}}}

{{{369=(4W+5)*W}}}

{{{369=4W^2+5W}}}

{{{4W^2+5W-369=0}}}

{{{(W - 9) (4 W + 41) = 0}}}

we need only positive solution:

{{{(W - 9) = 0}}}=>{{{highlight(W =9in)}}}


go to

{{{L=4W+5in}}}....eq1 plug in {{{W}}}

{{{L=4*9in+5in}}}

{{{L=36in}}}

{{{highlight(L=41in)}}}