Question 648540
let's the length e {{{L}}} and width {{{W}}}

the perimeter of a rectangle is {{{P=2(L+W)}}} is 94yd, and 

the area of the rectangle is {{{A=L*W}}} 90 yd^2.

given:

{{{P= 94yd}}}........1

{{{A=90yd^2}}}.......2
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{{{2(L+W)= 94yd}}}........1

{{{L*W=90yd^2}}}.......2
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{{{2(L+W)= 94}}}........1...solve for {{{L}}}

{{{L+W= 94/2}}}

{{{L= 47-W}}}.......substitute in 2

{{{(47-W)*W=90}}}.......2...solve for {{{W}}}

{{{47W-W*W=90}}}

{{{47W-W^2=90}}}

{{{0=W^2-47W+90}}} or

{{{W^2-47W+90=0}}}......use quadratic formula to find {{{W}}}

{{{W = (-(-47) +- sqrt((-47)^2-4*1*90 ))/(2*1) }}}

{{{W = (47 +- sqrt(2209-360 ))/2 }}}

{{{W = (47 +- sqrt(1849))/2 }}}

{{{W = (47 +- 43)/2 }}}...the roots are

{{{W = (47 + 43)/2 }}}

{{{W = 90/2 }}}

{{{highlight(W = 45) }}}

or

{{{W = (47 - 43)/2 }}}

{{{W = 4/2 }}}

{{{highlight(W = 2) }}}

now find {{{L}}}


{{{L= 47-W}}}..

{{{L= 47-45}}}

{{{highlight(L= 2)}}}

or

{{{L= 47-W}}}..

{{{L= 47-2}}}

{{{highlight(L= 45)}}}


one pair solutions is: {{{highlight(L= 2)}}} and {{{highlight(W = 45) }}} 

another pair is {{{highlight(L=45)}}} and {{{highlight(W = 2) }}}