Precalculus Examples

$6 - y + (2 x + 8) i = 7 - 4 i$

Step 1

Apply the distributive property.

$6 - y + 2 x i + 8 i = 7 - 4 i$

Step 2

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

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Step 2.1

Subtract $6$ from both sides of the equation.

$- y + 2 x i + 8 i = 7 - 4 i - 6$

Step 2.2

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

$- y + 8 i = 7 - 4 i - 6 - 2 x i$

Step 2.3

Subtract $8 i$ from both sides of the equation.

$- y = 7 - 4 i - 6 - 2 x i - 8 i$

Step 2.4

Subtract $6$ from $7$ .

$- y = 1 - 4 i - 2 x i - 8 i$

Step 2.5

Subtract $8 i$ from $- 4 i$ .

$- y = 1 - 2 x i - 12 i$

Step 3

Divide each term in $- y = 1 - 2 x i - 12 i$ by $- 1$ and simplify.

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Step 3.1

Divide each term in $- y = 1 - 2 x i - 12 i$ by $- 1$ .

$\frac{- y}{- 1} = \frac{1}{- 1} + \frac{- 2 x i}{- 1} + \frac{- 12 i}{- 1}$

Step 3.2

Simplify the left side.

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Step 3.2.1

Dividing two negative values results in a positive value.

$\frac{y}{1} = \frac{1}{- 1} + \frac{- 2 x i}{- 1} + \frac{- 12 i}{- 1}$

Step 3.2.2

Divide $y$ by $1$ .

$y = \frac{1}{- 1} + \frac{- 2 x i}{- 1} + \frac{- 12 i}{- 1}$

Step 3.3

Simplify the right side.

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Step 3.3.1

Simplify each term.

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Step 3.3.1.1

Divide $1$ by $- 1$ .

$y = - 1 + \frac{- 2 x i}{- 1} + \frac{- 12 i}{- 1}$

Step 3.3.1.2

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

$y = - 1 - 1 \cdot (- 2 x i) + \frac{- 12 i}{- 1}$

Step 3.3.1.3

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

$y = - 1 - (- 2 x i) + \frac{- 12 i}{- 1}$

Step 3.3.1.4

Multiply $- 2$ by $- 1$ .

$y = - 1 + 2 (x i) + \frac{- 12 i}{- 1}$

Step 3.3.1.5

Move the negative one from the denominator of $\frac{- 12 i}{- 1}$ .

$y = - 1 + 2 x i - 1 \cdot (- 12 i)$

Step 3.3.1.6

Rewrite $- 1 \cdot (- 12 i)$ as $- (- 12 i)$ .

$y = - 1 + 2 x i - (- 12 i)$

Step 3.3.1.7

Multiply $- 12$ by $- 1$ .

$y = - 1 + 2 x i + 12 i$