Splitting Sixty-Four, Homework Help?

Hey, i need some homework help please, i've found the answer but need to make sure i've done the right steps to get there. Here is the problem i must answer:





Sixty-Four can be split into four parts so that when 3 is added to the first part, 3 is subrtacted from the second part, the third part is multiplied by 3, and the fourth part is divided by 3, all four parts wil be equal. What are the four parts?





If you do help me, make sure that you show the steps on how to get where you're going, and explain how you got what, so i know if i made a mistake, thanks! I'd appreaciate the help!Splitting Sixty-Four, Homework Help?
1st equation:


(x+3) + (x-3) + (3X) + (X/3) = 64


X3 to get rid of fraction


3[(x+3) + (x-3) + (3X) + (X/3)] = 3*64


distribute the 3:


3x + 9 + 3x - 9 + 9x + x + 192


simplify:


16x = 192


Divide by 16


x = 12


then you reverse the steps:


1st # (x-3) (not x+3, b/c you reverse the order)= 9


2nd # (x+3) = 15


3rd # (x/3) = 4


4th # (3x) = 36Splitting Sixty-Four, Homework Help?
The ';catch';: No matter if you begin your equation by adding, subtracting, multiplying, or dividing, your ORIGINAL number will be 64. That's the ';common thread'; to the 4.