Well on this problem recall area of a rectangle is defined by:
A = lw our w = 1/6 and our l = (x + [tex] \frac{2}{3} [/tex]) we have let 'x' represent our original length.
[tex](x + \frac{2}{3} )( \frac{1}{6} )= \frac{11}{72} [/tex]
[tex] \frac{1}{6} x+ \frac{2}{18} = \frac{11}{72} [/tex]
12x + 8 = 11
12x = 11 - 8
12x = 3
x = [tex] \frac{3}{12} = \frac{1}{4} [/tex] So our original length is [tex] \frac{1}{4} [/tex]
