Solve for a 3a+(4b+2)i=9-6i

$3 a + (4 b + 2) i = 9 - 6 i$

Apply the distributive property.

$3 a + 4 b i + 2 i = 9 - 6 i$

Move all terms not containing $a$ to the right side of the equation.

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Subtract $4 b i$ from both sides of the equation.

$3 a + 2 i = 9 - 6 i - 4 b i$

Subtract $2 i$ from both sides of the equation.

$3 a = 9 - 6 i - 4 b i - 2 i$

Subtract $2 i$ from $- 6 i$ .

$3 a = 9 - 4 b i - 8 i$

Divide each term by $3$ and simplify.

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Divide each term in $3 a = 9 - 4 b i - 8 i$ by $3$ .

$\frac{3 a}{3} = \frac{9}{3} + \frac{- 4 b i}{3} + \frac{- 8 i}{3}$

Cancel the common factor of $3$ .

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Cancel the common factor.

$\frac{3 a}{3} = \frac{9}{3} + \frac{- 4 b i}{3} + \frac{- 8 i}{3}$

Divide $a$ by $1$ .

$a = \frac{9}{3} + \frac{- 4 b i}{3} + \frac{- 8 i}{3}$

Simplify each term.

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Divide $9$ by $3$ .

$a = 3 + \frac{- 4 b i}{3} + \frac{- 8 i}{3}$

Move the negative in front of the fraction.

$a = 3 - \frac{(4) b i}{3} + \frac{- 8 i}{3}$

Move the negative in front of the fraction.

$a = 3 - \frac{4 b i}{3} - \frac{8 i}{3}$

Solve for a 3a+(4b+2)i=9-6i

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