Solve for y -10x+12i=20+3yi

$- 10 x + 12 i = 20 + 3 y i$

Rewrite the equation as $20 + 3 y i = - 10 x + 12 i$ .

$20 + 3 y i = - 10 x + 12 i$

Subtract $20$ from both sides of the equation.

$3 y i = - 10 x + 12 i - 20$

Divide each term in $3 y i = - 10 x + 12 i - 20$ by $3 i$ and simplify.

Divide each term in $3 y i = - 10 x + 12 i - 20$ by $3 i$ .

$\frac{3 y i}{3 i} = \frac{- 10 x}{3 i} + \frac{12 i}{3 i} + \frac{- 20}{3 i}$

Simplify the left side.

Cancel the common factor of $3$ .

Cancel the common factor.

$\frac{3 y i}{3 i} = \frac{- 10 x}{3 i} + \frac{12 i}{3 i} + \frac{- 20}{3 i}$

Rewrite the expression.

$\frac{y i}{i} = \frac{- 10 x}{3 i} + \frac{12 i}{3 i} + \frac{- 20}{3 i}$

Cancel the common factor of $i$ .

Cancel the common factor.

$\frac{y i}{i} = \frac{- 10 x}{3 i} + \frac{12 i}{3 i} + \frac{- 20}{3 i}$

Divide $y$ by $1$ .

$y = \frac{- 10 x}{3 i} + \frac{12 i}{3 i} + \frac{- 20}{3 i}$

Simplify the right side.

Simplify each term.

Multiply the numerator and denominator of $\frac{- 10 x}{3 i}$ by the conjugate of $3 i$ to make the denominator real.

$y = \frac{- 10 x}{3 i} \cdot \frac{i}{i} + \frac{12 i}{3 i} + \frac{- 20}{3 i}$

Multiply.

Combine.

$y = \frac{- 10 x i}{3 i i} + \frac{12 i}{3 i} + \frac{- 20}{3 i}$

Simplify the denominator.

Add parentheses.

$y = \frac{- 10 x i}{3 (i i)} + \frac{12 i}{3 i} + \frac{- 20}{3 i}$

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

$y = \frac{- 10 x i}{3 (i^{1} i)} + \frac{12 i}{3 i} + \frac{- 20}{3 i}$

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

$y = \frac{- 10 x i}{3 (i^{1} i^{1})} + \frac{12 i}{3 i} + \frac{- 20}{3 i}$

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

$y = \frac{- 10 x i}{3 i^{1 + 1}} + \frac{12 i}{3 i} + \frac{- 20}{3 i}$

Add $1$ and $1$ .

$y = \frac{- 10 x i}{3 i^{2}} + \frac{12 i}{3 i} + \frac{- 20}{3 i}$

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

$y = \frac{- 10 x i}{3 \cdot - 1} + \frac{12 i}{3 i} + \frac{- 20}{3 i}$

Multiply $3$ by $- 1$ .

$y = \frac{- 10 x i}{- 3} + \frac{12 i}{3 i} + \frac{- 20}{3 i}$

Dividing two negative values results in a positive value.

$y = \frac{(10) x i}{3} + \frac{12 i}{3 i} + \frac{- 20}{3 i}$

Cancel the common factor of $12$ and $3$ .

Factor $3$ out of $12 i$ .

$y = \frac{10 x i}{3} + \frac{3 (4 i)}{3 i} + \frac{- 20}{3 i}$

Cancel the common factors.

Factor $3$ out of $3 i$ .

$y = \frac{10 x i}{3} + \frac{3 (4 i)}{3 (i)} + \frac{- 20}{3 i}$

Cancel the common factor.

$y = \frac{10 x i}{3} + \frac{3 (4 i)}{3 i} + \frac{- 20}{3 i}$

Rewrite the expression.

$y = \frac{10 x i}{3} + \frac{4 i}{i} + \frac{- 20}{3 i}$

Cancel the common factor of $i$ .

Cancel the common factor.

$y = \frac{10 x i}{3} + \frac{4 i}{i} + \frac{- 20}{3 i}$

Divide $4$ by $1$ .

$y = \frac{10 x i}{3} + 4 + \frac{- 20}{3 i}$

Multiply the numerator and denominator of $\frac{- 20}{3 i}$ by the conjugate of $3 i$ to make the denominator real.

$y = \frac{10 x i}{3} + 4 + \frac{- 20}{3 i} \cdot \frac{i}{i}$

Multiply.

Combine.

$y = \frac{10 x i}{3} + 4 + \frac{- 20 i}{3 i i}$

Simplify the denominator.

Add parentheses.

$y = \frac{10 x i}{3} + 4 + \frac{- 20 i}{3 (i i)}$

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

$y = \frac{10 x i}{3} + 4 + \frac{- 20 i}{3 (i^{1} i)}$

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

$y = \frac{10 x i}{3} + 4 + \frac{- 20 i}{3 (i^{1} i^{1})}$

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

$y = \frac{10 x i}{3} + 4 + \frac{- 20 i}{3 i^{1 + 1}}$

Add $1$ and $1$ .

$y = \frac{10 x i}{3} + 4 + \frac{- 20 i}{3 i^{2}}$

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

$y = \frac{10 x i}{3} + 4 + \frac{- 20 i}{3 \cdot - 1}$

Multiply $3$ by $- 1$ .

$y = \frac{10 x i}{3} + 4 + \frac{- 20 i}{- 3}$

Dividing two negative values results in a positive value.

$y = \frac{10 x i}{3} + 4 + \frac{20 i}{3}$

Solve for y -10x+12i=20+3yi

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