Question 1160885

A rectangular billboard sign has a length that is four yards longer than twice its width, x If the area of the sign is 30 square yards, which equation could be used to find the dimensions in yards of the sign?

if a billboard sign length is {{{4yd}}} longer than twice its width we have

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

if its area is {{{30yd^2}}}, we have

{{{L*W=30yd^2}}}........eq.2

substitute {{{L}}} from eq.1 in eq.2

{{{(2W+4yd)*W=30yd^2}}}............solve for {{{W}}}

{{{2W^2+4yd*W=30yd^2}}}

{{{2W^2+4yd*W-30yd^2=0}}}...simplify, divide by {{{2}}}

{{{W^2+2yd*W-15yd^2=0}}}........ factor completely

{{{W^2-3yd*W+5yd*W-15yd^2=0}}}

{{{((W^2-3yd*W)+(5yd*W-15yd^2))=0}}}

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

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


=> {{{(W - 3yd) = 0}}}->{{{W = 3yd}}}
=> {{{(W +5yd) = 0}}}->{{{W = -5yd}}}->disregard negative solution



go  to {{{L=2W+4yd}}} .........eq.1, substitute {{{W}}}


{{{L=2*3yd+4yd}}} 

{{{L=10yd}}} 


 so, the dimensions of the rectangular billboard sign are:

the length: {{{10yd}}} 

the width: {{{3yd}}}