Question 44059: solve each of the following systems by using either addition or substitution. If a unique solution does not exist, state whether the system is dependent or inconsistent.
10x+2y=7
y=-5x+3
Answer by venugopalramana(3286)   (Show Source):
You can put this solution on YOUR website! SEE THE FOLLOWING AND TRY.IF STILL IN DIFFICULTY PLEASE COME BACK
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IS the given system of equations consistent
MEANS WHETHER THERE IS ANY SET OF VALUES FOR X AND Y
WHICH SATISFIES BOTH THE GIVEN EQUATIONS.
and dependent,MEANS WHETHER ONE EQUATION CAN BE GOT
FROM THE OTHER BY SUITABLE AND PERMISSIBLE
TRANSFORMATION
consistent and independent,MEANS THAT ONE EQUATION
CANNOT BE OBTAINED FROM THE OTHER EQUATION BY BY ANY
SUITABLE AND PERMISSIBLE TRANSFORMATION .
or inconsistent.MEANS THERE ARE NO SET OF VALUES FOR X
AND Y WHICH SATISFY BOTH THE EQUATIONS.
3x-y=7.....I.......
6x-2Y=9..........II
what does these terms mean?NOW I THINK YOU KNOW WHAT
THE TERMS MEAN.THERE ARE TESTS FOR FINDING THESE
SITUATIONS.WE SHALL RESTRICT THE ANALYSIS TO YOUR
PROBLEM
WE FIND THAT ON L.H.S OF THE 2 EQNS.WE HAVE 3X-Y AND
6X-2Y WHICH ARE DIRECT MULTIPLES BY 2 .THAT IS WE GET
LHS OF EQN.2 BY MULTIPLYING THE EQN 1 WITH 2
WE GET EQN.I *2 ....6X-2Y=14
WHERE AS EQN.II SAYS 6X-2Y=9..SO IF WE TRY TO SATISFY
EQN.I ,EQN II WILL NOT BE SATISFIED AND IF WE TRY TO
SATISFY EQN II ,EQN.I WILL NOT BE SATISFIED.SO THERE
ARE NO VALUES OF X AND Y WHICH SATISFY BOTH
EQUATIONS.SO WE SAY THE GIVEN EQNS. ARE INCONSISTENT.
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Solve the following system by addition. If a unique
solution does not exist, state
whether the system is inconsistent or dependent.
x + 5y =
10...................................................I
-2x – 10y = -20.................................II
-2*EQN.I GIVES
-2X-10Y=-20....WHICH IS SAME AS EQN.II...
HENCE THEY ARE DEPENDENT EQUATIONS.
SO INEFFECT WE HAVE ONLY ONE EQN.TO SOLVE FOR 2
UNKOWNS.
HENCE WE SHALL HAVE INFINTE SOLUTIONS LIKE IN THIS
CASE
X=0....Y=2
X=10...Y=0...ETC..
IMPORTANT NOTE.
THEY HAVE INFINITE SOLUTIONS BUT NOT ANY AND EVERY
VALUE OF X AND Y.FOR EXAMPLE IF X=0 ,THEN Y HAS TO
EQUAL 2 ONLY NOT ANY OTHER VALUE.SO IT HAS INFINITE
SETS OF X AND Y PAIRS FOR ITS SOLUTION
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2x+y=4
3x-y=6
make a deteminant with coefficients of x (2,3)and
y(1,-1) in the 2 eqns.call it C.(Actually for a
determinant as you know ,the numbers are contained in
vertical bars at either end like |xx|,but in the
following the bars are omitted due to difficulty in
depiction.you may assume the bars are present)
C=matrix(2,2,2,1,3,-1)=2*(-1)-(1*3)=-5
..now use the constants (4,6)to replace coefficients
of x(2,3) in the above determinant C...call it CX..
CX=matrix(2,2,4,1,6,-1)=4*(-1)-1*6=-4-6=-10
..now use the constants (4,6)to replace coefficients
of y(1,-1) in the above determinant C...call it CY..
CY=matrix(2,2,2,4,3,6)=2*6-3*4=12=12=0
..now cramers rule says that
(x/CX)=(y/CY)=(1/C)..so we get
x/(-10)=y/0=1/-5
x=-10/-5=10/5=2
y=0/-5=0
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............... 3x+5y=17
................. x+3y=15
D=DETERMINANT OF COEFFICIENT
D =DET|3,5|
............|1,3|=3*3-1*5=9-5=4
DX=DET|17,5|
............|15,3|=17*3-5*15=51-75=-24
DY=DET|3,17|
.............|1,15|=3*15-17*1=45-17=28
X/DX=Y/DY=1/D
X=DX/D=-24/4=-6
Y=DY/D=28/4=7
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