Question 461475
The width of a rectangle {{{y}}} is {{{3}}} less than twice the length {{{x}}}. 

{{{y+3=2x}}}.....->..{{{y=2x-3}}}


If the area of the rectangle is {{{A=43ft^2}}}, 

which equation can be used to find the length, in feet?

{{{xy=43ft^2}}}

{{{x(2x-3)=43ft^2}}}

{{{2x^2-3x=43ft^2}}}

{{{2x^2-3x-43ft^2=0}}}....use quadratic formula to find {{{x}}}

{{{x = (-b +- sqrt( b^2-4*a*c ))/(2*a) }}}

{{{x = (-(-3) +- sqrt( (-3)^2-4*2*(-43) ))/(2*2) }}}

{{{x = (3 +- sqrt(9+344 ))/4 }}}

{{{x = (3 +- sqrt(353 ))/4 }}}

{{{x = (3 +18.788294228055935999045204838699)/4 }}}

{{{x = 21.79/4 }}}

{{{x = 5.45ft}}}


{{{y=2x-3}}}

{{{y=2(5.45)-3}}}

{{{y=10.9-3}}}

{{{y=7.9ft}}}


check:

{{{5.45ft*7.9ft=43ft^2}}}

{{{43.055ft^2=43ft^2}}}

{{{43ft^2=43ft^2}}}