Question 938590
Area = length x width = 1102 in^2
Perimeter = L + W + L + W = 2L + 2W = 134 in
          = 2(L+W)= 134 in
So:
   L+W=67 in and LxW=1102 in^2
Solve the first equation for L:
   (L+W)-W=67-W
         L=67-W
Substitute in the other equation and solve for W:
     (67-W)x W= 1102 in^2
     67W - W^2= 1102 in^2...................Add w^2 to both sides
 67W-W^2+(W^2)= 1102 in^2 + (w^2)...........Subtract 67W from both sides
       67W-67W= 1102 in^2 + W^2 - 67W.......Re-arrange right side
             0= W^2 - 67W + 1102 
             0= (W-38)(W-29)
Thus we have W-38=0 and w-29=0
 so length is 38 and width is 29
Check:
area = L x W=38 in x 29 in =1102 in^2, the given
perimeter = 2(38 in)+ 2(29 in)=76 + 58= 134 in, also given