Solve for y 8+(3x+y)i=2x-4i

$8 + (3 x + y) i = 2 x - 4 i$

Apply the distributive property.

$8 + 3 x i + y i = 2 x - 4 i$

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

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Subtract $8$ from both sides of the equation.

$3 x i + y i = 2 x - 4 i - 8$

Subtract $3 x i$ from both sides of the equation.

$y i = 2 x - 4 i - 8 - 3 x i$

Divide each term by $i$ and simplify.

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Divide each term in $y i = 2 x - 4 i - 8 - 3 x i$ by $i$ .

$\frac{y i}{i} = \frac{2 x}{i} + \frac{- 4 i}{i} + \frac{- 8}{i} + \frac{- 3 x i}{i}$

Cancel the common factor of $i$ .

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Cancel the common factor.

$\frac{y i}{i} = \frac{2 x}{i} + \frac{- 4 i}{i} + \frac{- 8}{i} + \frac{- 3 x i}{i}$

Divide $y$ by $1$ .

$y = \frac{2 x}{i} + \frac{- 4 i}{i} + \frac{- 8}{i} + \frac{- 3 x i}{i}$

Simplify each term.

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Multiply the numerator and denominator of $\frac{2 x}{i}$ by the conjugate of $i$ to make the denominator real.

$y = \frac{2 x}{i} \cdot \frac{i}{i} + \frac{- 4 i}{i} + \frac{- 8}{i} + \frac{- 3 x i}{i}$

Multiply.

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Combine.

$y = \frac{2 x i}{i i} + \frac{- 4 i}{i} + \frac{- 8}{i} + \frac{- 3 x i}{i}$

Simplify the denominator.

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Raise $i$ to the power of $1$ .

$y = \frac{2 x i}{i^{1} i} + \frac{- 4 i}{i} + \frac{- 8}{i} + \frac{- 3 x i}{i}$

Raise $i$ to the power of $1$ .

$y = \frac{2 x i}{i^{1} i^{1}} + \frac{- 4 i}{i} + \frac{- 8}{i} + \frac{- 3 x i}{i}$

Use the power rule $a^{m} a^{n} = a^{m + n}$ to combine exponents.

$y = \frac{2 x i}{i^{1 + 1}} + \frac{- 4 i}{i} + \frac{- 8}{i} + \frac{- 3 x i}{i}$

Add $1$ and $1$ .

$y = \frac{2 x i}{i^{2}} + \frac{- 4 i}{i} + \frac{- 8}{i} + \frac{- 3 x i}{i}$

Rewrite $i^{2}$ as $- 1$ .

$y = \frac{2 x i}{- 1} + \frac{- 4 i}{i} + \frac{- 8}{i} + \frac{- 3 x i}{i}$

Move the negative one from the denominator of $\frac{2 x i}{- 1}$ .

$y = - 1 \cdot (2 x i) + \frac{- 4 i}{i} + \frac{- 8}{i} + \frac{- 3 x i}{i}$

Rewrite $- 1 \cdot (2 x i)$ as $- (2 x i)$ .

$y = - (2 x i) + \frac{- 4 i}{i} + \frac{- 8}{i} + \frac{- 3 x i}{i}$

Multiply $2$ by $- 1$ .

$y = - 2 (x i) + \frac{- 4 i}{i} + \frac{- 8}{i} + \frac{- 3 x i}{i}$

Cancel the common factor of $i$ .

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Cancel the common factor.

$y = - 2 x i + \frac{- 4 i}{i} + \frac{- 8}{i} + \frac{- 3 x i}{i}$

Divide $- 4$ by $1$ .

$y = - 2 x i - 4 + \frac{- 8}{i} + \frac{- 3 x i}{i}$

Multiply the numerator and denominator of $\frac{- 8}{i}$ by the conjugate of $i$ to make the denominator real.

$y = - 2 x i - 4 + \frac{- 8}{i} \cdot \frac{i}{i} + \frac{- 3 x i}{i}$

Multiply.

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Combine.

$y = - 2 x i - 4 + \frac{- 8 i}{i i} + \frac{- 3 x i}{i}$

Simplify the denominator.

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Raise $i$ to the power of $1$ .

$y = - 2 x i - 4 + \frac{- 8 i}{i^{1} i} + \frac{- 3 x i}{i}$

Raise $i$ to the power of $1$ .

$y = - 2 x i - 4 + \frac{- 8 i}{i^{1} i^{1}} + \frac{- 3 x i}{i}$

Use the power rule $a^{m} a^{n} = a^{m + n}$ to combine exponents.

$y = - 2 x i - 4 + \frac{- 8 i}{i^{1 + 1}} + \frac{- 3 x i}{i}$

Add $1$ and $1$ .

$y = - 2 x i - 4 + \frac{- 8 i}{i^{2}} + \frac{- 3 x i}{i}$

Rewrite $i^{2}$ as $- 1$ .

$y = - 2 x i - 4 + \frac{- 8 i}{- 1} + \frac{- 3 x i}{i}$

Move the negative one from the denominator of $\frac{- 8 i}{- 1}$ .

$y = - 2 x i - 4 - 1 \cdot (- 8 i) + \frac{- 3 x i}{i}$

Rewrite $- 1 \cdot (- 8 i)$ as $- (- 8 i)$ .

$y = - 2 x i - 4 - (- 8 i) + \frac{- 3 x i}{i}$

Multiply $- 8$ by $- 1$ .

$y = - 2 x i - 4 + 8 i + \frac{- 3 x i}{i}$

Cancel the common factor of $i$ .

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Cancel the common factor.

$y = - 2 x i - 4 + 8 i + \frac{- 3 x i}{i}$

Divide $- 3 x$ by $1$ .

$y = - 2 x i - 4 + 8 i - 3 x$

Solve for y 8+(3x+y)i=2x-4i

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