ExplanationConvert this word problem into two equations with two variables. “The length is two more than twice the width” can be written as:
\(L = 2W + 2\)
Since the area is 40 and area is equal to length × width:
\(LW = 40\)
Since the first equation is already solved for L, plug (2W + 2) in for L into the second equation:
\((2W + 2)W = 40\)
\(2W^2 + 2W = 40\)
Since this is now a quadratic (there are both a W2 and a W term), get all terms on one side to set the expression equal to zero:
\(2W^2 + 2W - 40 = 0\)
Simplify as much as possible—in this case, divide the entire equation by 2—before trying to factor:
\(W^2 + W - 20 = 0\)
\((W - 4)(W + 5) = 0\)
W = 4 or –5
Since a width cannot be negative, the width is equal to 4. Since LW is equal to 40, the length must be 10. Now use the equation for perimeter to solve:
Perimeter = 2L + 2W
Perimeter = 2(10) + 2(4)
Perimeter = 28
Note that it might have been possible for you to puzzle out that the sides were 4 and 10 just by trying values. However, if you did this, you got lucky—no one said that the values even had to be integers!
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