Question 87249
           let the length &width of derek's rectangle be = l & b
            area of the rectangle  = 35 sq mm
           length & width of David's rectangle = (l+3)&(b+3)
           area of David's rectangle = 143 sq mm
            (l+3)(b+3)=143  lb+3l+3b+9 =143
                           35+3(l+b)+9 = 143
                            3(l+b)= 143-44
                            3(l+b)=99
                               l+b = 99/3 = 33          l = 33-b--------eq 1
                             l.b = 35   b.(33-b)=35  
                                         33b-bsquare = 35

                      b(square)-33b+35=0
                   use x=(-b+-sqrt(b^2-4ac)/2.a       b=33+-sqrt33^2-4.1.(-35)/2
                                                   b= 33+-sqrt1229/2
                                                   b= 33+-35.05/2                  
                                                  b = 68.05/2=34.03
                                              or b=-2.05/2= -1.03

                       l=34.03 if b=-1.03  and l=-1.03 if b=34.03

   length &width of David's rectangle = l+3=37.03  b=1.97  or vice versa