Question 972887
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W=original width; L=original length=W+4ft
W+3ft=new width; L+3ft=new length=W+7ft
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NewArea=new length x new width
{{{NewArea=(W+7ft)(W+3ft)}}}
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{{{NewArea=W^2+10W+21}}}ANSWER: This is the equation:Area={{{W^2+10W+21}}}
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CHECK: (The assigned width is arbitrary, just to check.  Any number will work.)
Let W=2 feet
L=W+4ft=6 feet
New length=L+3 feet=6 feet+3 feet=9 feet
New width=W+3 feet=2 feet+3 feet=5 feet
New Area=new length x new width=9 feet x 5 feet=45 square feet
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CHECK New Equation:
{{{NewArea=W^2+10W+21}}}
{{{NewArea=(2ft)^2+10ft(2ft)+21ft^2}}}
{{{NewArea=4ft^2+20ft^2+21ft^2}}}
{{{NewArea=45ft^2}}} Same as using (new length) x (new width)