Question 909431


A rectangular parking lot has length {{{L}}} that is {{{7yd}}} less than {{{twice}}} its width {{{W}}}. 

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

If the area of the land is {{{A=130yd^2}}}, then we know that 

{{{L*W=130yd^2}}} ... plug in {{{L=2W-7}}} 

{{{(2W-7)*W=130}}}

{{{2W^2-7W=130}}}

{{{2W^2-7W-130=0}}} ...write {{{-7W}}} as {{{13W-20W}}}

{{{W^2+13W-20W-130 = 0}}} ...group

{{{(W^2+13W)-(20W+130) = 0}}}

{{{W(W+13)-10(2W+13) = 0}}}

{{{(W-10)(2W+13) = 0}}}

solutions:

if {{{W-10 = 0}}} => {{{W=10}}}

if {{{W+13 = 0}}} => {{{W=-13}}} => we don't need this solution since the width  cannot be negative

so, {{{highlight(W=10ft)}}}

now find the length {{{L=2W-7}}} 

{{{L=2*10-7}}} 

{{{L=20-7}}} 

{{{highlight(L=13ft)}}} 

so, the dimensions of the land are: {{{highlight(L=13ft)}}} by {{{highlight(W=10ft)}}}