Question 1028221
Let w=width of first rectangle, and a=area. Original length, l, is w+8;
w(w+8)=a
w^2+8w=a
3w^2+24w=3a
The second rectangle is;
(w+8)(w+16)=3a
w^2+24w+128=3w^2+24w
2w^2=128
w^2=64
w=8
The original rectangle was 8 ft in width; 16 feet length!!!!!!!!!!