Question 1129448


le's the length of a rectangle be {{{L}}} and the  width {{{W}}}

if the length is {{{4}}} yards more than {{{twice}}} its width, we have

{{{L=2W+4}}} ....eq.1

if the area is {{{70}}} square yards, we have

{{{L*W=70}}}..........substitute {{{L}}} from eq.1

{{{(2W+4)*W=70}}}

{{{2W^2+4W=70}}}...simplify

{{{W^2+2W=35}}}

{{{W^2+2W-35=0}}}....factor

{{{W^2+7W-5W-35=0}}}

{{{(W^2+7W)-(5W+35)=0}}}

{{{W(W+7)-5(W+7)=0}}}

{{{(W - 5) (W + 7) = 0}}}

since we are looking for width, we need only positive solution


{{{(W - 5) = 0}}}=>{{{W=5}}}...............the width

go back to eq.1

{{{L=2W+4}}} ....eq.1
{{{L=2*5+4}}} 

{{{L=14}}} ....the length