SOLUTION: {{{((3+i)/(2-i))(a+bi)=1}}} In the equation above, a and b are constants. If {{{i=sqrt(-1)}}}. what is the value of a?
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Question 1043213: In the equation above, a and b are constants. If . what is the value of a? Found 2 solutions by stanbon, ikleyn:Answer by stanbon(75887) (Show Source):
You can put this solution on YOUR website! ((3+i)/(2-i))(a+bi)=1 In the equation above, a and b are constants. If i=sqrt(-1). what is the value of a?
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(6-3i+2i+1)(a+bi) = 1
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(7-i)(a+bi) = 1 + 0i
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7a+b + (7b-a)i = 1 + 0i
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7a + b = 1
-a + 7b = 0
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Modify for elimination::
7a + b = 1
-7a + 49b = 0
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Add and solve for "b"::
50b = 1
b = 1/50
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Solve for "a":
7a + b = 1
7a = (49/50)
a = 7/50
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Cheers,
Stan H.
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You can put this solution on YOUR website! . In the equation above, a and b are constants. If . what is the value of a?
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If = then (a+bi) is the number reciprocal to :
a + bi = = ,
and all we need to do is to rid of the denominator in the fraction .
For it, multiply the fraction by . Since the last fraction is 1, multiplication will not change our number .
People also say "let us multiply the numerator and the denominator of an original fraction by (3-i) ").
So, we have
= =
and since = -1,
= = = 0.5 - 0.5*i.
Thus we got a + bi = 0.5 - 0.5*i.
It implies a = 0.5, b = -0.5. The problem is solved. Your answer is: a = 0.5, b = -0.5.
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