SOLUTION: I can't figure out how to solve this equation 6x-3y=1 -12x+6y=-2 I need to solve using a matrix but I can't figure out how to get the answer in the back of the book. It says th

Question 39698This question is from textbook College Algebra
: I can't figure out how to solve this equation
6x-3y=1
-12x+6y=-2 I need to solve using a matrix but I can't figure out how to get the answer in the back of the book. It says the answer is (3y+1/6,y) This question is from textbook College Algebra

Answer by venugopalramana(3286) (Show Source): You can put this solution on YOUR website!
I can't figure out how to solve this equation
6x-3y=1
-12x+6y=-2 I need to solve using a matrix but I can't figure out how to get the answer in the back of the book. It says the answer is (3y+1/6,y)
THESE ARE DEPENDENT EQNS..HENCE YOU MIGHT BE FACING THE DIFFICULTY
HOPE YOU KNOW THE METHOD
AUGMENTED MATRIX IS
....COEF.OF X.....COEF.OF Y.....CONSTANT......UNIT MATRIX
............6.......-3..............1...........1..............0
...........-12..... .6.............-2...........0..............1

WE HAVE TO CONVERT IT NOW..
1.NEW R1=OLD R1/6
............1........-3/6...........1/6.........1/6.............0
............-12........6.............-2..........0..............1
2.NEW R2 =OLD R2+12*R1
............1.........-1/2...........1/6.........1/6.............0
............0..........0.............0............2..............1
NOW WE CANNOPT MAKE SECOND ROW ,SECOND COLUMN ELEMENT AS 1 ANY MORE AS IS NORMALLY DONE IN CONSISTENT AND INDEPENDENT EQN.
THE PRESENCE OF ZEROS IN ALL 3 COLUMNS IN SECOND ROW INDICATES THAT THE EQN.GIVEN ARE DEPENDENT .SO WE HAVE ONLY ONE INDEPENDENT EQN.IN 2 UNKNOWNS .HENCE WE SHALL NOT HAVE A UNIQUE SOLUTION.WE GET AN INFINITE SET OF ANSWERS .THEY ARE GIVEN BY FIRST ROW WHICH TELLS US THAT
1X-Y/2=1/6
SO TO GET ANY SET OF SOLUTION PUT X ANY VALUE AND FIND Y OR VICEVERSA.
SUPPOSE WE PUT X=0...THEN -Y/2=1/6...OR....Y=-2/6=-1/3...
SO SOLUTION SET IS (0,-1/3)
OR PUT X=1/6....THEN Y=0..
SO SOLUTION SET IS (*1/6,0)....ETC....
..SAME WAY WITH Y
PUT Y=0...THEN X=1/6....
SO SOLUTION SET IS (1/6,0)
IN GENERAL IF WE TAKE Y=Y..THEN 1X=Y/2+1/6 = (3Y+1)/6
SO THE SOLUTION SET IS
{(3Y+1)/6,Y}..THIS IS WHAT IS GIVEN IN YOUR BOOK.
IT MEANS PUT ANY VALUE FOR Y IN THIS SET AND THE RESULTANT NUMBERS WILL GIVE YOU A SOLUTION FOR THE GIVEN EQNS.
BUT YOU NEED NOT GIVE THIS ONLY AS AN ANSWER.FOR EX. YOU CAN PUT X=X AND THEN
X-Y/2=1/6...SO....Y/2=X-1/6....OR....Y=2X-1/3=(6X-1)/3...HENCE THE ANSWER COULD ALSO BE GIVEN AS {X,(6X-1)/3}...GOT IT?BUT YOU HAVE TO GIVE A GENERAL SOLUTION LIKE THIS AND NOT JUST 1 SET OF VALUES LIKE (0,-1/3)
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