Question 55197
SEE THE FOLLOWING EXAMPLE AND TRY...IF STILL IN DIFFICULTY PLEASE COME BACK.
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I really don't understand this problem. Can you please help me?
3(1-3x)+2=4x+14........OK
3-9x+2=4x+14................GOOD
so then add or subtract from both sides?
3-9x+9x+2=13+14......HERE YOU GOT CONFUSED..THE STEPS INVOLVED ARE
1.REMOVE BRACKETS ......THIS YOU HAVE DONE.
2.TRANSFER ALL UNKNOWN TERMS TO ONE SIDE.... USUALLY L.H.S. AND KNOWN TERMS TO ONE SIDE ...USUALLY R.H.S.
3.THIS TRANSFER YOU CAN THINK OF AND DO AS FOLLOWS....
EVERY TERM HAS A + OR - SIGN INFRONT OF IT.WHEN YOU TRANSFER THAT TERM FROM ONE SIDE TO ANOTHER ,+ BECOMES - AND - BECOMES PLUS.
THIS IS BECAUSE...AS YOU WERE TRYING TO SAY 'ADD OR SUBTRACT FROM BOTH SIDES'
AN EQUATION IS UNALTERED IF WE ADD OR SUBTRACT SAME QUANTITY FROM BOTH SIDES OF THE EQN.SO IF WE WANT TO REMOVE +4X ABOVE FROM R.H.S.WE HAVE TO ADD -4X TO IT.SO TO BALANCE IT AND KEEP THE VALIDITY OF THE EQN.UNALTERED,WE ADD -4X TO L.H.S.ALSO.THEN ON R.H.S ,WE GET +4X-4X=0...THAT IS +4X DIAPPEARS AS WE WANTED.BUT -4X REMAINS ON L.H.S.....THIS IS THE FINAL EFFECT OF ADD AND SUBTRACT WHICH IS SUMMARISED FIRST ABOVE.
HENCE WE GET HERE
-9X-4X = 14-2-3 
4.NOW SIMPLIFY BOTH SIDES COMBINING ALL UNKNOWNS ON L.H.S AND ALL KNOWNS ON R.H.S
WE GET HERE 
-13X=9
5.NOW TRANSFER THE COEFFICIENT OF UNKNOWN ON L.H.S.TO R.H.S.
APPLYING THE SAME RULE & LOGIC MENTIONED ABOVE,HERE * WILL BECOME / AND / BECOMES *
HERE WE GET
X = -9/13....
THIS IS THE ANSWER.
6.IF YOU HAVE TIME AND WOULD LIKE TO DO....CHECK BACK THE ANSWER FOR CONFIRMATION BY SUBSTITUTION..
3{1-3*(-9/13)}+2=4*(-9/13)+ 14
3{1+27/13)+2=-36/13 +14
3*40/13 +2=(-36+14*13)/13
120/13 +2= 146/13
146/13 = 146/13.......OK