text: # SOLUTION: I just don't understand these -2-4(2y-1)=(6-2y)3
  
  Algebra ->  Algebra  -> Functions -> SOLUTION: I just don't understand these -2-4(2y-1)=(6-2y)3      Log On
  
   Ad: Algebra Solved!™: algebra software solves algebra homework problems with step-by-step help! Ad: Algebrator™ solves your algebra problems and provides step-by-step explanations!
  
   Algebra: Functions, Domain, NOT graphing Solvers Lessons Answers archive Quiz In Depth
  
   Question 51028: I just don't understand these -2-4(2y-1)=(6-2y)3Answer by junior403(76)   (Show Source): You can put this solution on YOUR website!OK, no problem... Lets just take these one step at a time. -2-4(2y-1)=(6-2y)3 What if we just work on one side of the equation at a time, that might help. So all we have is the left side: -2-4(2y-1) If we follow the order of operations we know that we should work with the parenthesis first using the distributive property, forget about the -2 for now. So we have... -4 X 2y and -4 X -1 this becomes -8y and -4 or -8y -4 ok so far? now if we bring back the -2 we have... -2 -8y -4 Does this make sense? Then we can combine like terms (the -2 and the -4) -8y -6 Ok? Now we put that aside and work on the right side of the equation... (6-2y)3 or we could look at this like 3(6-2y) So now we have... 3 X 6 and 3 X -2y this becomes 18 and -6y or 18 - 6y ok so far? now that we have re-written the equation lets bring it back together... -8y -6 = 18 - 6y does this make sense? now we need to solve for y... we can combine like terms again accross the equals sign. In order to do that we can add 6 to both sides and we add 6y from both sides to make it look like this... -8y - 6 + 6 + 6y = 18 + 6 - 6y + 6y the 6's will cancel on the left side and the 6y's will cancel on the right. -8y + 6y = 18 + 6 now we can combine the terms... -2y = 24 ok so far? now to completely isolate the y variable we can divide both sides by -2... the -2's cancel on the left side and we can reduce the right fraction to -12 so our final answer is... y = -12 I hope this helps. Please let me know if you have any more questions! Good Luck. Alan