Finite Math Examples

$(\frac{- x - 1 + 4 i}{1 + 2 i - i x}) i (2 - i) + 1 = 4 i - x (1 - \frac{- x - 1 + 4 i}{1 + 2 i - i x} i)$

Step 1

Since $x$ is on the right side of the equation, switch the sides so it is on the left side of the equation.

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

Step 2

Simplify $4 i - x (1 - \frac{- x - 1 + 4 i}{1 + 2 i - i x} \cdot i)$ .

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Step 2.1

Simplify each term.

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Step 2.1.1

Simplify each term.

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Step 2.1.1.1

Reorder terms.

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

Step 2.1.1.2

Combine $i$ and $\frac{- x - 1 + 4 i}{- i x + 1 + 2 i}$ .

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

Step 2.1.1.3

Apply the distributive property.

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

Step 2.1.1.4

Simplify.

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Step 2.1.1.4.1

Move $- 1$ to the left of $i$ .

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

Step 2.1.1.4.2

Multiply $i (4 i)$ .

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Step 2.1.1.4.2.1

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

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

Step 2.1.1.4.2.2

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

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

Step 2.1.1.4.2.3

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

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

Step 2.1.1.4.2.4

Add $1$ and $1$ .

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

Step 2.1.1.5

Simplify each term.

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Step 2.1.1.5.1

Rewrite $- 1 i$ as $- i$ .

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

Step 2.1.1.5.2

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

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

Step 2.1.1.5.3

Multiply $4$ by $- 1$ .

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

Step 2.1.1.6

Reorder factors in $i (- x) - i - 4$ .

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

Step 2.1.1.7

Reorder terms.

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

Step 2.1.2

Write $1$ as a fraction with a common denominator.

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

Step 2.1.3

Combine the numerators over the common denominator.

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

Step 2.1.4

Simplify the numerator.

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Step 2.1.4.1

Apply the distributive property.

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

Step 2.1.4.2

Simplify.

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Step 2.1.4.2.1

Multiply $- (- i x)$ .

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Step 2.1.4.2.1.1

Multiply $- 1$ by $- 1$ .

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

Step 2.1.4.2.1.2

Multiply $x$ by $1$ .

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

Step 2.1.4.2.2

Multiply $- 1$ by $- 4$ .

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

Step 2.1.4.2.3

Multiply $- 1$ by $- 1$ .

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

Step 2.1.4.3

Multiply $i$ by $1$ .

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

Step 2.1.4.4

Add $- i x$ and $x i$ .

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Step 2.1.4.4.1

Move $i$ .

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

Step 2.1.4.4.2

Add $- 1 x i$ and $x i$ .

$4 i - x \frac{0 + 1 + 2 i + 4 + i}{- i x + 1 + 2 i} = \frac{- x - 1 + 4 i}{1 + 2 i - i x} \cdot (i (2 - i)) + 1$

Step 2.1.4.5

Add $0$ and $1$ .

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

Step 2.1.4.6

Add $1$ and $4$ .

$4 i - x \frac{5 + 2 i + i}{- i x + 1 + 2 i} = \frac{- x - 1 + 4 i}{1 + 2 i - i x} \cdot (i (2 - i)) + 1$

Step 2.1.4.7

Add $2 i$ and $i$ .

$4 i - x \frac{5 + 3 i}{- i x + 1 + 2 i} = \frac{- x - 1 + 4 i}{1 + 2 i - i x} \cdot (i (2 - i)) + 1$

Step 2.1.5

Combine $\frac{5 + 3 i}{- i x + 1 + 2 i}$ and $x$ .

$4 i - \frac{(5 + 3 i) x}{- i x + 1 + 2 i} = \frac{- x - 1 + 4 i}{1 + 2 i - i x} \cdot (i (2 - i)) + 1$

Step 2.2

To write $4 i$ as a fraction with a common denominator, multiply by $\frac{- i x + 1 + 2 i}{- i x + 1 + 2 i}$ .

$\frac{4 i (- i x + 1 + 2 i)}{- i x + 1 + 2 i} - \frac{(5 + 3 i) x}{- i x + 1 + 2 i} = \frac{- x - 1 + 4 i}{1 + 2 i - i x} \cdot (i (2 - i)) + 1$

Step 2.3

Combine the numerators over the common denominator.

$\frac{4 i (- i x + 1 + 2 i) - (5 + 3 i) x}{- i x + 1 + 2 i} = \frac{- x - 1 + 4 i}{1 + 2 i - i x} \cdot (i (2 - i)) + 1$

Step 2.4

Factor $- 1$ out of $- i x$ .

$\frac{4 i (- i x + 1 + 2 i) - (5 + 3 i) x}{- (i x) + 1 + 2 i} = \frac{- x - 1 + 4 i}{1 + 2 i - i x} \cdot (i (2 - i)) + 1$

Step 2.5

Rewrite $1$ as $- 1 (- 1)$ .

$\frac{4 i (- i x + 1 + 2 i) - (5 + 3 i) x}{- (i x) - 1 (- 1) + 2 i} = \frac{- x - 1 + 4 i}{1 + 2 i - i x} \cdot (i (2 - i)) + 1$

Step 2.6

Factor $- 1$ out of $- (i x) - 1 (- 1)$ .

$\frac{4 i (- i x + 1 + 2 i) - (5 + 3 i) x}{- (i x - 1) + 2 i} = \frac{- x - 1 + 4 i}{1 + 2 i - i x} \cdot (i (2 - i)) + 1$

Step 2.7

Factor $- 1$ out of $2 i$ .

$\frac{4 i (- i x + 1 + 2 i) - (5 + 3 i) x}{- (i x - 1) - (- 2 i)} = \frac{- x - 1 + 4 i}{1 + 2 i - i x} \cdot (i (2 - i)) + 1$

Step 2.8

Factor $- 1$ out of $- (i x - 1) - (- 2 i)$ .

$\frac{4 i (- i x + 1 + 2 i) - (5 + 3 i) x}{- (i x - 1 - 2 i)} = \frac{- x - 1 + 4 i}{1 + 2 i - i x} \cdot (i (2 - i)) + 1$

Step 2.9

Rewrite negatives.

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Step 2.9.1

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

$\frac{4 i (- i x + 1 + 2 i) - (5 + 3 i) x}{- 1 (i x - 1 - 2 i)} = \frac{- x - 1 + 4 i}{1 + 2 i - i x} \cdot (i (2 - i)) + 1$

Step 2.9.2

Move the negative in front of the fraction.

$- \frac{4 i (- i x + 1 + 2 i) - (5 + 3 i) x}{i x - 1 - 2 i} = \frac{- x - 1 + 4 i}{1 + 2 i - i x} \cdot (i (2 - i)) + 1$

Step 2.9.3

Reorder factors in $- \frac{4 i (- i x + 1 + 2 i) - (5 + 3 i) x}{i x - 1 - 2 i}$ .

$- \frac{4 i (- i x + 1 + 2 i) - x (5 + 3 i)}{i x - 1 - 2 i} = \frac{- x - 1 + 4 i}{1 + 2 i - i x} \cdot (i (2 - i)) + 1$

Step 3

Simplify $\frac{- x - 1 + 4 i}{1 + 2 i - i x} \cdot (i (2 - i)) + 1$ .

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Step 3.1

Simplify each term.

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Step 3.1.1

Reorder terms.

$- \frac{4 i (- i x + 1 + 2 i) - x (5 + 3 i)}{i x - 1 - 2 i} = \frac{- x - 1 + 4 i}{- i x + 1 + 2 i} \cdot (i (2 - i)) + 1$

Step 3.1.2

Apply the distributive property.

$- \frac{4 i (- i x + 1 + 2 i) - x (5 + 3 i)}{i x - 1 - 2 i} = \frac{- x - 1 + 4 i}{- i x + 1 + 2 i} \cdot (i \cdot 2 + i (- i)) + 1$

Step 3.1.3

Move $2$ to the left of $i$ .

$- \frac{4 i (- i x + 1 + 2 i) - x (5 + 3 i)}{i x - 1 - 2 i} = \frac{- x - 1 + 4 i}{- i x + 1 + 2 i} \cdot (2 \cdot i + i (- i)) + 1$

Step 3.1.4

Multiply $i (- i)$ .

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Step 3.1.4.1

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

$- \frac{4 i (- i x + 1 + 2 i) - x (5 + 3 i)}{i x - 1 - 2 i} = \frac{- x - 1 + 4 i}{- i x + 1 + 2 i} \cdot (2 \cdot i - (i^{1} i)) + 1$

Step 3.1.4.2

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

$- \frac{4 i (- i x + 1 + 2 i) - x (5 + 3 i)}{i x - 1 - 2 i} = \frac{- x - 1 + 4 i}{- i x + 1 + 2 i} \cdot (2 \cdot i - (i^{1} i^{1})) + 1$

Step 3.1.4.3

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

$- \frac{4 i (- i x + 1 + 2 i) - x (5 + 3 i)}{i x - 1 - 2 i} = \frac{- x - 1 + 4 i}{- i x + 1 + 2 i} \cdot (2 \cdot i - i^{1 + 1}) + 1$

Step 3.1.4.4

Add $1$ and $1$ .

$- \frac{4 i (- i x + 1 + 2 i) - x (5 + 3 i)}{i x - 1 - 2 i} = \frac{- x - 1 + 4 i}{- i x + 1 + 2 i} \cdot (2 \cdot i - i^{2}) + 1$

Step 3.1.5

Simplify each term.

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Step 3.1.5.1

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

$- \frac{4 i (- i x + 1 + 2 i) - x (5 + 3 i)}{i x - 1 - 2 i} = \frac{- x - 1 + 4 i}{- i x + 1 + 2 i} \cdot (2 i - - 1) + 1$

Step 3.1.5.2

Multiply $- 1$ by $- 1$ .

$- \frac{4 i (- i x + 1 + 2 i) - x (5 + 3 i)}{i x - 1 - 2 i} = \frac{- x - 1 + 4 i}{- i x + 1 + 2 i} \cdot (2 i + 1) + 1$

Step 3.1.6

Multiply $\frac{- x - 1 + 4 i}{- i x + 1 + 2 i}$ by $2 i + 1$ .

$- \frac{4 i (- i x + 1 + 2 i) - x (5 + 3 i)}{i x - 1 - 2 i} = \frac{(- x - 1 + 4 i) (2 i + 1)}{- i x + 1 + 2 i} + 1$

Step 3.2

Simplify terms.

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Step 3.2.1

Write $1$ as a fraction with a common denominator.

$- \frac{4 i (- i x + 1 + 2 i) - x (5 + 3 i)}{i x - 1 - 2 i} = \frac{(- x - 1 + 4 i) (2 i + 1)}{- i x + 1 + 2 i} + \frac{- i x + 1 + 2 i}{- i x + 1 + 2 i}$

Step 3.2.2

Combine the numerators over the common denominator.

$- \frac{4 i (- i x + 1 + 2 i) - x (5 + 3 i)}{i x - 1 - 2 i} = \frac{(- x - 1 + 4 i) (2 i + 1) - i x + 1 + 2 i}{- i x + 1 + 2 i}$

Step 3.2.3

Factor $- 1$ out of $- i x$ .

$- \frac{4 i (- i x + 1 + 2 i) - x (5 + 3 i)}{i x - 1 - 2 i} = \frac{(- x - 1 + 4 i) (2 i + 1) - i x + 1 + 2 i}{- (i x) + 1 + 2 i}$

Step 3.2.4

Rewrite $1$ as $- 1 (- 1)$ .

$- \frac{4 i (- i x + 1 + 2 i) - x (5 + 3 i)}{i x - 1 - 2 i} = \frac{(- x - 1 + 4 i) (2 i + 1) - i x + 1 + 2 i}{- (i x) - 1 (- 1) + 2 i}$

Step 3.2.5

Factor $- 1$ out of $- (i x) - 1 (- 1)$ .

$- \frac{4 i (- i x + 1 + 2 i) - x (5 + 3 i)}{i x - 1 - 2 i} = \frac{(- x - 1 + 4 i) (2 i + 1) - i x + 1 + 2 i}{- (i x - 1) + 2 i}$

Step 3.2.6

Factor $- 1$ out of $2 i$ .

$- \frac{4 i (- i x + 1 + 2 i) - x (5 + 3 i)}{i x - 1 - 2 i} = \frac{(- x - 1 + 4 i) (2 i + 1) - i x + 1 + 2 i}{- (i x - 1) - (- 2 i)}$

Step 3.2.7

Factor $- 1$ out of $- (i x - 1) - (- 2 i)$ .

$- \frac{4 i (- i x + 1 + 2 i) - x (5 + 3 i)}{i x - 1 - 2 i} = \frac{(- x - 1 + 4 i) (2 i + 1) - i x + 1 + 2 i}{- (i x - 1 - 2 i)}$

Step 3.2.8

Simplify the expression.

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Step 3.2.8.1

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

$- \frac{4 i (- i x + 1 + 2 i) - x (5 + 3 i)}{i x - 1 - 2 i} = \frac{(- x - 1 + 4 i) (2 i + 1) - i x + 1 + 2 i}{- 1 (i x - 1 - 2 i)}$

Step 3.2.8.2

Move the negative in front of the fraction.

$- \frac{4 i (- i x + 1 + 2 i) - x (5 + 3 i)}{i x - 1 - 2 i} = - \frac{(- x - 1 + 4 i) (2 i + 1) - i x + 1 + 2 i}{i x - 1 - 2 i}$

Step 3.2.8.3

Reorder $2 i$ and $1$ .

$- \frac{4 i (- i x + 1 + 2 i) - x (5 + 3 i)}{i x - 1 - 2 i} = - \frac{(- x - 1 + 4 i) (1 + 2 i) - i x + 1 + 2 i}{i x - 1 - 2 i}$

Step 4

Move all terms containing $x$ to the left side of the equation.

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Step 4.1

Add $\frac{(- x - 1 + 4 i) (1 + 2 i) - i x + 1 + 2 i}{i x - 1 - 2 i}$ to both sides of the equation.

$- \frac{4 i (- i x + 1 + 2 i) - x (5 + 3 i)}{i x - 1 - 2 i} + \frac{(- x - 1 + 4 i) (1 + 2 i) - i x + 1 + 2 i}{i x - 1 - 2 i} = 0$

Step 4.2

Combine the numerators over the common denominator.

$\frac{- (4 i (- i x + 1 + 2 i) - x (5 + 3 i)) + (- x - 1 + 4 i) (1 + 2 i) - i x + 1 + 2 i}{i x - 1 - 2 i} = 0$

Step 4.3

Simplify each term.

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Step 4.3.1

Simplify each term.

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Step 4.3.1.1

Apply the distributive property.

$\frac{- (4 i (- i x) + 4 i \cdot 1 + 4 i (2 i) - x (5 + 3 i)) + (- x - 1 + 4 i) (1 + 2 i) - i x + 1 + 2 i}{i x - 1 - 2 i} = 0$

Step 4.3.1.2

Simplify.

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Step 4.3.1.2.1

Multiply $4 i (- i x)$ .

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Step 4.3.1.2.1.1

Multiply $- 1$ by $4$ .

$\frac{- (- 4 i (i x) + 4 i \cdot 1 + 4 i (2 i) - x (5 + 3 i)) + (- x - 1 + 4 i) (1 + 2 i) - i x + 1 + 2 i}{i x - 1 - 2 i} = 0$

Step 4.3.1.2.1.2

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

$\frac{- (- 4 (i^{1} i) x + 4 i \cdot 1 + 4 i (2 i) - x (5 + 3 i)) + (- x - 1 + 4 i) (1 + 2 i) - i x + 1 + 2 i}{i x - 1 - 2 i} = 0$

Step 4.3.1.2.1.3

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

$\frac{- (- 4 (i^{1} i^{1}) x + 4 i \cdot 1 + 4 i (2 i) - x (5 + 3 i)) + (- x - 1 + 4 i) (1 + 2 i) - i x + 1 + 2 i}{i x - 1 - 2 i} = 0$

Step 4.3.1.2.1.4

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

$\frac{- (- 4 i^{1 + 1} x + 4 i \cdot 1 + 4 i (2 i) - x (5 + 3 i)) + (- x - 1 + 4 i) (1 + 2 i) - i x + 1 + 2 i}{i x - 1 - 2 i} = 0$

Step 4.3.1.2.1.5

Add $1$ and $1$ .

$\frac{- (- 4 i^{2} x + 4 i \cdot 1 + 4 i (2 i) - x (5 + 3 i)) + (- x - 1 + 4 i) (1 + 2 i) - i x + 1 + 2 i}{i x - 1 - 2 i} = 0$

Step 4.3.1.2.2

Multiply $4$ by $1$ .

$\frac{- (- 4 i^{2} x + 4 i + 4 i (2 i) - x (5 + 3 i)) + (- x - 1 + 4 i) (1 + 2 i) - i x + 1 + 2 i}{i x - 1 - 2 i} = 0$

Step 4.3.1.2.3

Multiply $4 i (2 i)$ .

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Step 4.3.1.2.3.1

Multiply $2$ by $4$ .

$\frac{- (- 4 i^{2} x + 4 i + 8 i i - x (5 + 3 i)) + (- x - 1 + 4 i) (1 + 2 i) - i x + 1 + 2 i}{i x - 1 - 2 i} = 0$

Step 4.3.1.2.3.2

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

$\frac{- (- 4 i^{2} x + 4 i + 8 (i^{1} i) - x (5 + 3 i)) + (- x - 1 + 4 i) (1 + 2 i) - i x + 1 + 2 i}{i x - 1 - 2 i} = 0$

Step 4.3.1.2.3.3

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

$\frac{- (- 4 i^{2} x + 4 i + 8 (i^{1} i^{1}) - x (5 + 3 i)) + (- x - 1 + 4 i) (1 + 2 i) - i x + 1 + 2 i}{i x - 1 - 2 i} = 0$

Step 4.3.1.2.3.4

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

$\frac{- (- 4 i^{2} x + 4 i + 8 i^{1 + 1} - x (5 + 3 i)) + (- x - 1 + 4 i) (1 + 2 i) - i x + 1 + 2 i}{i x - 1 - 2 i} = 0$

Step 4.3.1.2.3.5

Add $1$ and $1$ .

$\frac{- (- 4 i^{2} x + 4 i + 8 i^{2} - x (5 + 3 i)) + (- x - 1 + 4 i) (1 + 2 i) - i x + 1 + 2 i}{i x - 1 - 2 i} = 0$

Step 4.3.1.3

Simplify each term.

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Step 4.3.1.3.1

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

$\frac{- (- 4 \cdot - 1 x + 4 i + 8 i^{2} - x (5 + 3 i)) + (- x - 1 + 4 i) (1 + 2 i) - i x + 1 + 2 i}{i x - 1 - 2 i} = 0$

Step 4.3.1.3.2

Multiply $- 4$ by $- 1$ .

$\frac{- (4 x + 4 i + 8 i^{2} - x (5 + 3 i)) + (- x - 1 + 4 i) (1 + 2 i) - i x + 1 + 2 i}{i x - 1 - 2 i} = 0$

Step 4.3.1.3.3

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

$\frac{- (4 x + 4 i + 8 \cdot - 1 - x (5 + 3 i)) + (- x - 1 + 4 i) (1 + 2 i) - i x + 1 + 2 i}{i x - 1 - 2 i} = 0$

Step 4.3.1.3.4

Multiply $8$ by $- 1$ .

$\frac{- (4 x + 4 i - 8 - x (5 + 3 i)) + (- x - 1 + 4 i) (1 + 2 i) - i x + 1 + 2 i}{i x - 1 - 2 i} = 0$

Step 4.3.1.4

Apply the distributive property.

$\frac{- (4 x + 4 i - 8 - x \cdot 5 - x (3 i)) + (- x - 1 + 4 i) (1 + 2 i) - i x + 1 + 2 i}{i x - 1 - 2 i} = 0$

Step 4.3.1.5

Multiply $5$ by $- 1$ .

$\frac{- (4 x + 4 i - 8 - 5 x - x (3 i)) + (- x - 1 + 4 i) (1 + 2 i) - i x + 1 + 2 i}{i x - 1 - 2 i} = 0$

Step 4.3.1.6

Multiply $3$ by $- 1$ .

$\frac{- (4 x + 4 i - 8 - 5 x - 3 x i) + (- x - 1 + 4 i) (1 + 2 i) - i x + 1 + 2 i}{i x - 1 - 2 i} = 0$

Step 4.3.2

Subtract $5 x$ from $4 x$ .

$\frac{- (- x + 4 i - 8 - 3 x i) + (- x - 1 + 4 i) (1 + 2 i) - i x + 1 + 2 i}{i x - 1 - 2 i} = 0$

Step 4.3.3

Apply the distributive property.

$\frac{- - x - (4 i) - - 8 - (- 3 x i) + (- x - 1 + 4 i) (1 + 2 i) - i x + 1 + 2 i}{i x - 1 - 2 i} = 0$

Step 4.3.4

Simplify.

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Step 4.3.4.1

Multiply $- - x$ .

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Step 4.3.4.1.1

Multiply $- 1$ by $- 1$ .

$\frac{1 x - (4 i) - - 8 - (- 3 x i) + (- x - 1 + 4 i) (1 + 2 i) - i x + 1 + 2 i}{i x - 1 - 2 i} = 0$

Step 4.3.4.1.2

Multiply $x$ by $1$ .

$\frac{x - (4 i) - - 8 - (- 3 x i) + (- x - 1 + 4 i) (1 + 2 i) - i x + 1 + 2 i}{i x - 1 - 2 i} = 0$

Step 4.3.4.2

Multiply $4$ by $- 1$ .

$\frac{x - 4 i - - 8 - (- 3 x i) + (- x - 1 + 4 i) (1 + 2 i) - i x + 1 + 2 i}{i x - 1 - 2 i} = 0$

Step 4.3.4.3

Multiply $- 1$ by $- 8$ .

$\frac{x - 4 i + 8 - (- 3 x i) + (- x - 1 + 4 i) (1 + 2 i) - i x + 1 + 2 i}{i x - 1 - 2 i} = 0$

Step 4.3.4.4

Multiply $- 3$ by $- 1$ .

$\frac{x - 4 i + 8 + 3 (x i) + (- x - 1 + 4 i) (1 + 2 i) - i x + 1 + 2 i}{i x - 1 - 2 i} = 0$

Step 4.3.5

Expand $(- x - 1 + 4 i) (1 + 2 i)$ by multiplying each term in the first expression by each term in the second expression.

$\frac{x - 4 i + 8 + 3 x i - x \cdot 1 - x (2 i) - 1 \cdot 1 - 1 (2 i) + 4 i \cdot 1 + 4 i (2 i) - i x + 1 + 2 i}{i x - 1 - 2 i} = 0$

Step 4.3.6

Simplify each term.

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Step 4.3.6.1

Multiply $- 1$ by $1$ .

$\frac{x - 4 i + 8 + 3 x i - x - x (2 i) - 1 \cdot 1 - 1 (2 i) + 4 i \cdot 1 + 4 i (2 i) - i x + 1 + 2 i}{i x - 1 - 2 i} = 0$

Step 4.3.6.2

Multiply $2$ by $- 1$ .

$\frac{x - 4 i + 8 + 3 x i - x - 2 x i - 1 \cdot 1 - 1 (2 i) + 4 i \cdot 1 + 4 i (2 i) - i x + 1 + 2 i}{i x - 1 - 2 i} = 0$

Step 4.3.6.3

Multiply $- 1$ by $1$ .

$\frac{x - 4 i + 8 + 3 x i - x - 2 x i - 1 - 1 (2 i) + 4 i \cdot 1 + 4 i (2 i) - i x + 1 + 2 i}{i x - 1 - 2 i} = 0$

Step 4.3.6.4

Multiply $2$ by $- 1$ .

$\frac{x - 4 i + 8 + 3 x i - x - 2 x i - 1 - 2 i + 4 i \cdot 1 + 4 i (2 i) - i x + 1 + 2 i}{i x - 1 - 2 i} = 0$

Step 4.3.6.5

Multiply $4$ by $1$ .

$\frac{x - 4 i + 8 + 3 x i - x - 2 x i - 1 - 2 i + 4 i + 4 i (2 i) - i x + 1 + 2 i}{i x - 1 - 2 i} = 0$

Step 4.3.6.6

Multiply $4 i (2 i)$ .

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Step 4.3.6.6.1

Multiply $2$ by $4$ .

$\frac{x - 4 i + 8 + 3 x i - x - 2 x i - 1 - 2 i + 4 i + 8 i i - i x + 1 + 2 i}{i x - 1 - 2 i} = 0$

Step 4.3.6.6.2

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

$\frac{x - 4 i + 8 + 3 x i - x - 2 x i - 1 - 2 i + 4 i + 8 (i^{1} i) - i x + 1 + 2 i}{i x - 1 - 2 i} = 0$

Step 4.3.6.6.3

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

$\frac{x - 4 i + 8 + 3 x i - x - 2 x i - 1 - 2 i + 4 i + 8 (i^{1} i^{1}) - i x + 1 + 2 i}{i x - 1 - 2 i} = 0$

Step 4.3.6.6.4

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

$\frac{x - 4 i + 8 + 3 x i - x - 2 x i - 1 - 2 i + 4 i + 8 i^{1 + 1} - i x + 1 + 2 i}{i x - 1 - 2 i} = 0$

Step 4.3.6.6.5

Add $1$ and $1$ .

$\frac{x - 4 i + 8 + 3 x i - x - 2 x i - 1 - 2 i + 4 i + 8 i^{2} - i x + 1 + 2 i}{i x - 1 - 2 i} = 0$

Step 4.3.6.7

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

$\frac{x - 4 i + 8 + 3 x i - x - 2 x i - 1 - 2 i + 4 i + 8 \cdot - 1 - i x + 1 + 2 i}{i x - 1 - 2 i} = 0$

Step 4.3.6.8

Multiply $8$ by $- 1$ .

$\frac{x - 4 i + 8 + 3 x i - x - 2 x i - 1 - 2 i + 4 i - 8 - i x + 1 + 2 i}{i x - 1 - 2 i} = 0$

Step 4.3.7

Subtract $8$ from $- 1$ .

$\frac{x - 4 i + 8 + 3 x i - x - 2 x i - 9 - 2 i + 4 i - i x + 1 + 2 i}{i x - 1 - 2 i} = 0$

Step 4.3.8

Add $- 2 i$ and $4 i$ .

$\frac{x - 4 i + 8 + 3 x i - x - 2 x i - 9 + 2 i - i x + 1 + 2 i}{i x - 1 - 2 i} = 0$

Step 4.4

Combine the opposite terms in $x - 4 i + 8 + 3 x i - x - 2 x i - 9 + 2 i - i x + 1 + 2 i$ .

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Step 4.4.1

Subtract $x$ from $x$ .

$\frac{0 - 4 i + 8 + 3 x i - 2 x i - 9 + 2 i - i x + 1 + 2 i}{i x - 1 - 2 i} = 0$

Step 4.4.2

Subtract $4 i$ from $0$ .

$\frac{- 4 i + 8 + 3 x i - 2 x i - 9 + 2 i - i x + 1 + 2 i}{i x - 1 - 2 i} = 0$

Step 4.5

Add $- 4 i$ and $2 i$ .

$\frac{8 + 3 x i - 2 x i - 9 - 2 i - i x + 1 + 2 i}{i x - 1 - 2 i} = 0$

Step 4.6

Combine the opposite terms in $8 + 3 x i - 2 x i - 9 - 2 i - i x + 1 + 2 i$ .

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Step 4.6.1

Add $- 2 i$ and $2 i$ .

$\frac{8 + 3 x i - 2 x i - 9 - i x + 1 + 0}{i x - 1 - 2 i} = 0$

Step 4.6.2

Add $8 + 3 x i - 2 x i - 9 - i x + 1$ and $0$ .

$\frac{8 + 3 x i - 2 x i - 9 - i x + 1}{i x - 1 - 2 i} = 0$

Step 4.7

Subtract $9$ from $8$ .

$\frac{3 x i - 2 x i - 1 - i x + 1}{i x - 1 - 2 i} = 0$

Step 4.8

Combine the opposite terms in $3 x i - 2 x i - 1 - i x + 1$ .

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Step 4.8.1

Add $- 1$ and $1$ .

$\frac{3 x i - 2 x i - i x + 0}{i x - 1 - 2 i} = 0$

Step 4.8.2

Add $3 x i - 2 x i - i x$ and $0$ .

$\frac{3 x i - 2 x i - i x}{i x - 1 - 2 i} = 0$

Step 4.9

Subtract $2 x i$ from $3 x i$ .

$\frac{1 x i - i x}{i x - 1 - 2 i} = 0$

Step 4.10

Multiply $x$ by $1$ .

$\frac{x i - i x}{i x - 1 - 2 i} = 0$

Step 4.11

Combine the opposite terms in $x i - i x$ .

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Step 4.11.1

Reorder the factors in the terms $x i$ and $- i x$ .

$\frac{i x - i x}{i x - 1 - 2 i} = 0$

Step 4.11.2

Subtract $i x$ from $i x$ .

$\frac{0}{i x - 1 - 2 i} = 0$

Step 4.12

Divide $0$ by $i x - 1 - 2 i$ .

$0 = 0$

Step 5

Since $0 = 0$ , the equation will always be true.

Always true