Question

a rectangle fish pond is 8 feet longer than it is wide. a wooden walk 2ft wide is placed around the pond. the area covered by the pond and the walk is 160 ft squared greater than the area covered by the pond alone. what are the dimensions of the pond?

Answer 1

L = 8 + w ;
l * w + 160 = ( l + 2 ) * ( w + 2 ) => l * w + 160 = l * w + 2 * l + 2 * w  + 4 => 156 = 2 * ( l + w ) => 78 = l + w; but, l = 8 + w => 78 = 8 + 2 * w => w = 70 / 2 => w = 35 feet; l = 8 + 35 => l = 43 feet.

Answer 2

$the\ fish\ pond:\\A_p=x\cdot (x+8)=x^2+8x\\\\the\ area\ covered\ by\ the\ walk:\\A_{w}=(x+2\cdot2)\cdot(x+8+2\cdot2)=(x+4)(x+12)=x^2+16x+48\\\\A_w=A_p+160\ \ \ \Rightarrow\ \ \ \ x^2+16x+48=x^2+8x+160\\\\16x-8x=160\ \ \ \Rightarrow\ \ \ 8x=160\ \ \ \Rightarrow\ \ \ x=20\\\\Ans.\ the\ width\ of\ the\ fish\ pond\ is\ 20\ ft\ and\ the\ length\ is\ 28\ ft$