Question 694264
15 yd^2 is the given area
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w = 3l - 4 ==> width given
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knowing that wl=a, substitute into the equation
l(3l-4) = 15
{{{3l^2 - 4l = 15}}}
{{{3l^2 - 4l - 15 = 0}}}
(3l+5)(l-3) = 0
3l+5=0
3l=-5
l = {{{-5/3}}}==> cannot have negative dimensions, so
l-3=0
l = 3
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If l=3, then
w = 3(3) - 4 = 9 - 4 = 5
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Dimensions are
3yd by 5yd
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