Question

Find the real value of a and b: a) (3a-6)+2ib=-6b+(6+a)i​

Answer 1

Answer:

$3a - 6 = 6b...(i) \\ 2b = 6 + a....(ii)$

Both eq has no any solution so given complex can't be equal

Answer 2

SOLUTION

TO DETERMINE

The real value of a and b :

(3a-6) + 2ib = - 6b + (6+a)i

EVALUATION

Here the given equation is

(3a-6) + 2ib = - 6b + (6+a)i

Comparing real part and Imaginary part in both sides we get

3a - 6 = - 6b

⇒ 3a + 6b = 6

⇒ a + 2b = 2 - - - - - (1)

Also

2b = 6 + a

⇒ a - 2b = - 6 - - - - - (2)

Equation 1 + Equation 2 gives

2a = - 4

⇒ a = - 2

From Equation 1 we get

- 2 + 2b = 2

⇒ 2b = 4

⇒ b = 2

FINAL ANSWER

Hence the required values are a = - 2 , b = 2

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