Question 54436: I need help. Solve the following system by using either addition or substitution. If a unique solution does not exist, state whether the system is dependent or inconsistent.
10x + 2y = 7 and y = -5x + 3
Found 2 solutions by Edwin McCravy, venugopalramana:
Answer by Edwin McCravy(20055)   (Show Source):
You can put this solution on YOUR website! I need help. Solve the following system by using
either addition or substitution. If a unique
solution does not exist, state whether the system
is dependent or inconsistent.
By substitution
10x + 2y = 7
y = -5x + 3
The second equation is already solved for y. So replace y
in the first equation by (-5x + 3)
10x + 2y = 7
10x + 2(-5x + 3) = 7
Solve that for x if possible
10x + 2(-5x + 3) = 7
10x - 10x + 9 = 7
9 = 7
Oh, oh! The unknowns cancelled out
and only a false numerical equation
9 = 7 resulted. So there can be no
solution. This is called an
inconsistent system. It has no
solution.
[If it had come out a true numerical
equation, such as 9 = 9 or 7 = 7,
then it would have been a dependent
system].
Edwin
Answer by venugopalramana(3286)  (Show Source):
You can put this solution on YOUR website! SEE THE FOLLOWING EXAMPLES AND TRY IF STILL IN DIFFICULTY PLEASE COME BACK
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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
ANOTHER EXAMPLE IS
X+Y=1........................1
2X+2Y=2..............................2
HERE AGIN EQN.2IS JUST A MULTIPLICATION OF EQN.1 WITH 2 ON BOTH LHS AND RHS
THE GERAL SOLUTION FORMULA IS Y=1-X...OR (X,1-X)IS A
SOLUTION SET..LIKE (1,0),(2,-1) ETC....
THUS DEPENDENT EQNS.HAVE LHS AND RHS IN THE SAME PRPORTION OR COMBINATION IN DIFFERENT EQNS.
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AS THE NAME IMPLIES AN INCONSISTENT SET OF EQUATIONS
HAS NO SOLUTION AS THEY ARE INCONSISTENT.FOR EXAMPLE
X+Y=2.....................1
2X+2Y=3.....................2
IF WE TRY TO SATISFY EQN.1,WE CANT SATISFY EQN.2 AND VICEVERSA...AS THE LHS OF BOTH EQNS.ARE INPROPORTION OF 1:2,WHERE AS RHS IS IN THE RATIO OF 2:3
HENCE THERE IS NO QUESTION OF GETTING A
GENERAL SOLUTION FOR THAT BY ANY METHOD .
SO INCONSISTENT EQNS HAVE LHS AS MULTIPLES OR COMBINATIONS IN ONE PROPORTION WHILE RHS IS IN DIFFERENT PROPORTION IN DIFFERENT EQNS.
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FINALLY CONSISTENT AND INDEPENDENT EQNS.HAVE UNIQUE
SOLUTION..FOR EXAMPLE
X+Y=2...................1
X-Y=0..........................2
EQN.1 +EQN.2 GIVES
2X=2...OR...X=1
SUBSTITUTING IN EQN.1 ,WE GET Y=1.THUS THEY
HAVE ONE UNIQUE SOLUTION X=1 AND Y=1
HERE LHS OF DIFFERENT EQNS.IS NOT A DIRECT MULTIPLE OR COMBINATIONS OF OTHER EQNS.
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