Question 44751: equations with 3 linear variables
6x-4y+5z=31
5x+2y+2z=13
x+y+z=2
should you multiply the 2nd equation by 2?
show work
Answer by venugopalramana(3286)   (Show Source):
You can put this solution on YOUR website! OK..THE PROBLEM,WE FACE IN ANSWERING IS THAT WE ARE NOT AWARE OF YOUR COURSE OF STUDY AND BACKGROUND.THE SOLUTION I GAVE IS BY GAUSS JORDAN ELIMINATION METHOD USED IN MATRICES AS WANTED BY SOME STUDENTS.IF YOU ARE NOT TAUGHT THAT ,WE SHALL DO BY NORMAL ELIMINATION.
equations with 3 linear variables
6x-4y+5z=31.......................................I
5x+2y+2z=13...........................II
x+y+z=2.......................III
EQN.II - 2*EQN.III
5X+2Y+2Z-2X-2Y-2Z=13-2*2=9
3X=9
X=9/3=3
EQN.I+4*EQN.III
6X-4Y+5Z+4X+4Y+4Z=31+4*2=39
10X+9Z=39...BUT...X=3..SO
10*3+9Z=39
9Z=39-30=9
Z=9/9=1
PUTTING IN EQN.III
3+Y+1=2
Y=2-4=-2
HOPE YOU UNDERSTOOD.
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SEE THE FOLLOWING AND TRY.IF STILL IN DIFFICULTY PLEASE COME BACK
YOU MAY MULTIPLY EQN.II WITH 2 AND ADD TO EQN.I TO ELIMINATE Y
x - y + z = 2
-x + y + z = 4
-x + z = 2
augmented matrix is
1.....................-1.......................1.............................2
-1.....................1.......................1.............................4
-1.....................0.......................1.............................2
nr2=r2+r1 and nr3=r3+r1
1..................-1.................1.............2
1-1=0............1-1=0............1+1=2..........2+4=6
1-1=0...........-1+0=-1...........1+1=2..........2+2=4
nr2=r2/2....nr3=r3-r2
1....................-1...............1..............2
0/2=0.............0/2=0..............2/2=1........6/2=3
0-0=0............-1-0=-1..............2-2=0.......4-6=-2
nr1=r1-r3-r2.......nr2=2r2........nr3=-r3
1-0-0=1............-1+1-0=0.........1-0-1=0.......2+2-3=1
0.....................0...................1............3
0.....................1...................0.............2
exchange r2 and r3
1....0........0.......1
0....1........0.......2
0....0........1.......3
he
nce x=1......y=2.....z=3
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