Question 146769
Let x=width and y=length



Since "The length of a rectangle is five less than twice the width", this means that {{{y=2x-5}}}. Also, because "The area is 42 square yards", this tells us that the area equation is {{{A=x*y}}} which becomes {{{42=x*y}}}



{{{42=x*y}}} Start with the second equation



{{{42=x*(2x-5)}}} Plug in {{{y=2x-5}}}



{{{42=2x^2-5x}}} Distribute



{{{42-2x^2+5x=0}}} Get all terms to the left side.



{{{-2x^2+5x+42=0}}} Combine like terms.



Notice we have a quadratic equation in the form of {{{ax^2+bx+c}}} where {{{a=-2}}}, {{{b=5}}}, and {{{c=42}}}



Let's use the quadratic formula to solve for x



{{{x = (-b +- sqrt( b^2-4ac ))/(2a)}}} Start with the quadratic formula



{{{x = (-(5) +- sqrt( (5)^2-4(-2)(42) ))/(2(-2))}}} Plug in  {{{a=-2}}}, {{{b=5}}}, and {{{c=42}}}



{{{x = (-5 +- sqrt( 25-4(-2)(42) ))/(2(-2))}}} Square {{{5}}} to get {{{25}}}. 



{{{x = (-5 +- sqrt( 25--336 ))/(2(-2))}}} Multiply {{{4(-2)(42)}}} to get {{{-336}}}



{{{x = (-5 +- sqrt( 25+336 ))/(2(-2))}}} Rewrite {{{sqrt(25--336)}}} as {{{sqrt(25+336)}}}



{{{x = (-5 +- sqrt( 361 ))/(2(-2))}}} Add {{{25}}} to {{{336}}} to get {{{361}}}



{{{x = (-5 +- sqrt( 361 ))/(-4)}}} Multiply {{{2}}} and {{{-2}}} to get {{{-4}}}. 



{{{x = (-5 +- 19)/(-4)}}} Take the square root of {{{361}}} to get {{{19}}}. 



{{{x = (-5 + 19)/(-4)}}} or {{{x = (-5 - 19)/(-4)}}} Break up the expression. 



{{{x = (14)/(-4)}}} or {{{x =  (-24)/(-4)}}} Combine like terms. 



{{{x = -7/2}}} or {{{x = 6}}} Simplify. 



So the widths are {{{x = -7/2}}} or {{{x = 6}}} 




However, since a negative width is not possible, this means that the only answer for the width is {{{x = 6}}}

  

{{{y=2x-5}}} Go back to the first equation



{{{y=2(6)-5}}} Plug in {{{x = 6}}}



{{{y=7}}} Simplify



So the width is 6 yards and the length is 7 lards