<PRE><B>Question 24733</B>
Before beginning this problem, there is a bit of a trick to it.  Notice that the length of the rectangle exceeds the width by 2 yards, but the perimeter is in FEET.  It might be a good idea to change the 2 yards to 6 feet.


Let x = width
x+6 = length

2(W) + 2(L) = Perimeter
2(x) + 2(x+6) = 88
2x + 2x + 12 = 88
4x + 12 = 88
4x +12 - 12 = 88-12
4x = 76


Divide both sides by 4:
{{{(4x)/4 = 76/4}}}
{{{x= 19}}} ft = width
{{{x + 6 = 25 }}} feet = length


Check:  
2(W) + 2(L) 
2(19) + 2(25)
38 + 50 = 88 feet 
It checks!


R^2 at SCC
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