Question 140973
First and formost Perimeter of rectangle(P)=2(Length+Width)---------- eqn 1.
But, there's a condition such being the Length 2 more than 3 times the Width.
Therefore, the Length =2+(3)(Width) 
Going back eqn 1 and substituting the values:
600=2(2+3W+W)
600=2(2+4W)
600=4+8W
W=74.50
Solving for L by reversing eqn 1.  becomes L=(P-2W)/2=(600-2*74.50)/2
L=225.50
Thank you,
Jojo