Precalculus Examples

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

Step 1

Simplify $(\frac{3 + 2 i}{2 - 3 i} + \frac{5 - i}{2 + 3 i}) \cdot \frac{a}{b}$ .

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Step 1.1

Simplify each term.

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Step 1.1.1

Multiply the numerator and denominator of $\frac{3 + 2 i}{2 - 3 i}$ by the conjugate of $2 - 3 i$ to make the denominator real.

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

Step 1.1.2

Multiply.

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Step 1.1.2.1

Combine.

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

Step 1.1.2.2

Simplify the numerator.

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Step 1.1.2.2.1

Expand $(3 + 2 i) (2 + 3 i)$ using the FOIL Method.

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Step 1.1.2.2.1.1

Apply the distributive property.

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

Step 1.1.2.2.1.2

Apply the distributive property.

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

Step 1.1.2.2.1.3

Apply the distributive property.

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

Step 1.1.2.2.2

Simplify and combine like terms.

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Step 1.1.2.2.2.1

Simplify each term.

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Step 1.1.2.2.2.1.1

Multiply $3$ by $2$ .

$(\frac{6 + 3 (3 i) + 2 i \cdot 2 + 2 i (3 i)}{(2 - 3 i) (2 + 3 i)} + \frac{5 - i}{2 + 3 i}) \cdot \frac{a}{b} = 1$

Step 1.1.2.2.2.1.2

Multiply $3$ by $3$ .

$(\frac{6 + 9 i + 2 i \cdot 2 + 2 i (3 i)}{(2 - 3 i) (2 + 3 i)} + \frac{5 - i}{2 + 3 i}) \cdot \frac{a}{b} = 1$

Step 1.1.2.2.2.1.3

Multiply $2$ by $2$ .

$(\frac{6 + 9 i + 4 i + 2 i (3 i)}{(2 - 3 i) (2 + 3 i)} + \frac{5 - i}{2 + 3 i}) \cdot \frac{a}{b} = 1$

Step 1.1.2.2.2.1.4

Multiply $2 i (3 i)$ .

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Step 1.1.2.2.2.1.4.1

Multiply $3$ by $2$ .

$(\frac{6 + 9 i + 4 i + 6 i i}{(2 - 3 i) (2 + 3 i)} + \frac{5 - i}{2 + 3 i}) \cdot \frac{a}{b} = 1$

Step 1.1.2.2.2.1.4.2

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

$(\frac{6 + 9 i + 4 i + 6 (i^{1} i)}{(2 - 3 i) (2 + 3 i)} + \frac{5 - i}{2 + 3 i}) \cdot \frac{a}{b} = 1$

Step 1.1.2.2.2.1.4.3

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

$(\frac{6 + 9 i + 4 i + 6 (i^{1} i^{1})}{(2 - 3 i) (2 + 3 i)} + \frac{5 - i}{2 + 3 i}) \cdot \frac{a}{b} = 1$

Step 1.1.2.2.2.1.4.4

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

$(\frac{6 + 9 i + 4 i + 6 i^{1 + 1}}{(2 - 3 i) (2 + 3 i)} + \frac{5 - i}{2 + 3 i}) \cdot \frac{a}{b} = 1$

Step 1.1.2.2.2.1.4.5

Add $1$ and $1$ .

$(\frac{6 + 9 i + 4 i + 6 i^{2}}{(2 - 3 i) (2 + 3 i)} + \frac{5 - i}{2 + 3 i}) \cdot \frac{a}{b} = 1$

Step 1.1.2.2.2.1.5

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

$(\frac{6 + 9 i + 4 i + 6 \cdot - 1}{(2 - 3 i) (2 + 3 i)} + \frac{5 - i}{2 + 3 i}) \cdot \frac{a}{b} = 1$

Step 1.1.2.2.2.1.6

Multiply $6$ by $- 1$ .

$(\frac{6 + 9 i + 4 i - 6}{(2 - 3 i) (2 + 3 i)} + \frac{5 - i}{2 + 3 i}) \cdot \frac{a}{b} = 1$

Step 1.1.2.2.2.2

Subtract $6$ from $6$ .

$(\frac{0 + 9 i + 4 i}{(2 - 3 i) (2 + 3 i)} + \frac{5 - i}{2 + 3 i}) \cdot \frac{a}{b} = 1$

Step 1.1.2.2.2.3

Add $0$ and $9 i$ .

$(\frac{9 i + 4 i}{(2 - 3 i) (2 + 3 i)} + \frac{5 - i}{2 + 3 i}) \cdot \frac{a}{b} = 1$

Step 1.1.2.2.2.4

Add $9 i$ and $4 i$ .

$(\frac{13 i}{(2 - 3 i) (2 + 3 i)} + \frac{5 - i}{2 + 3 i}) \cdot \frac{a}{b} = 1$

Step 1.1.2.3

Simplify the denominator.

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Step 1.1.2.3.1

Expand $(2 - 3 i) (2 + 3 i)$ using the FOIL Method.

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Step 1.1.2.3.1.1

Apply the distributive property.

$(\frac{13 i}{2 (2 + 3 i) - 3 i (2 + 3 i)} + \frac{5 - i}{2 + 3 i}) \cdot \frac{a}{b} = 1$

Step 1.1.2.3.1.2

Apply the distributive property.

$(\frac{13 i}{2 \cdot 2 + 2 (3 i) - 3 i (2 + 3 i)} + \frac{5 - i}{2 + 3 i}) \cdot \frac{a}{b} = 1$

Step 1.1.2.3.1.3

Apply the distributive property.

$(\frac{13 i}{2 \cdot 2 + 2 (3 i) - 3 i \cdot 2 - 3 i (3 i)} + \frac{5 - i}{2 + 3 i}) \cdot \frac{a}{b} = 1$

Step 1.1.2.3.2

Simplify.

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Step 1.1.2.3.2.1

Multiply $2$ by $2$ .

$(\frac{13 i}{4 + 2 (3 i) - 3 i \cdot 2 - 3 i (3 i)} + \frac{5 - i}{2 + 3 i}) \cdot \frac{a}{b} = 1$

Step 1.1.2.3.2.2

Multiply $3$ by $2$ .

$(\frac{13 i}{4 + 6 i - 3 i \cdot 2 - 3 i (3 i)} + \frac{5 - i}{2 + 3 i}) \cdot \frac{a}{b} = 1$

Step 1.1.2.3.2.3

Multiply $2$ by $- 3$ .

$(\frac{13 i}{4 + 6 i - 6 i - 3 i (3 i)} + \frac{5 - i}{2 + 3 i}) \cdot \frac{a}{b} = 1$

Step 1.1.2.3.2.4

Multiply $3$ by $- 3$ .

$(\frac{13 i}{4 + 6 i - 6 i - 9 i i} + \frac{5 - i}{2 + 3 i}) \cdot \frac{a}{b} = 1$

Step 1.1.2.3.2.5

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

$(\frac{13 i}{4 + 6 i - 6 i - 9 (i^{1} i)} + \frac{5 - i}{2 + 3 i}) \cdot \frac{a}{b} = 1$

Step 1.1.2.3.2.6

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

$(\frac{13 i}{4 + 6 i - 6 i - 9 (i^{1} i^{1})} + \frac{5 - i}{2 + 3 i}) \cdot \frac{a}{b} = 1$

Step 1.1.2.3.2.7

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

$(\frac{13 i}{4 + 6 i - 6 i - 9 i^{1 + 1}} + \frac{5 - i}{2 + 3 i}) \cdot \frac{a}{b} = 1$

Step 1.1.2.3.2.8

Add $1$ and $1$ .

$(\frac{13 i}{4 + 6 i - 6 i - 9 i^{2}} + \frac{5 - i}{2 + 3 i}) \cdot \frac{a}{b} = 1$

Step 1.1.2.3.2.9

Subtract $6 i$ from $6 i$ .

$(\frac{13 i}{4 + 0 - 9 i^{2}} + \frac{5 - i}{2 + 3 i}) \cdot \frac{a}{b} = 1$

Step 1.1.2.3.2.10

Add $4$ and $0$ .

$(\frac{13 i}{4 - 9 i^{2}} + \frac{5 - i}{2 + 3 i}) \cdot \frac{a}{b} = 1$

Step 1.1.2.3.3

Simplify each term.

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Step 1.1.2.3.3.1

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

$(\frac{13 i}{4 - 9 \cdot - 1} + \frac{5 - i}{2 + 3 i}) \cdot \frac{a}{b} = 1$

Step 1.1.2.3.3.2

Multiply $- 9$ by $- 1$ .

$(\frac{13 i}{4 + 9} + \frac{5 - i}{2 + 3 i}) \cdot \frac{a}{b} = 1$

Step 1.1.2.3.4

Add $4$ and $9$ .

$(\frac{13 i}{13} + \frac{5 - i}{2 + 3 i}) \cdot \frac{a}{b} = 1$

Step 1.1.3

Cancel the common factor of $13$ .

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Step 1.1.3.1

Cancel the common factor.

$(\frac{13 i}{13} + \frac{5 - i}{2 + 3 i}) \cdot \frac{a}{b} = 1$

Step 1.1.3.2

Divide $i$ by $1$ .

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

Step 1.1.4

Multiply the numerator and denominator of $\frac{5 - i}{2 + 3 i}$ by the conjugate of $2 + 3 i$ to make the denominator real.

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

Step 1.1.5

Multiply.

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Step 1.1.5.1

Combine.

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

Step 1.1.5.2

Simplify the numerator.

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Step 1.1.5.2.1

Expand $(5 - i) (2 - 3 i)$ using the FOIL Method.

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Step 1.1.5.2.1.1

Apply the distributive property.

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

Step 1.1.5.2.1.2

Apply the distributive property.

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

Step 1.1.5.2.1.3

Apply the distributive property.

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

Step 1.1.5.2.2

Simplify and combine like terms.

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Step 1.1.5.2.2.1

Simplify each term.

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Step 1.1.5.2.2.1.1

Multiply $5$ by $2$ .

$(i + \frac{10 + 5 (- 3 i) - i \cdot 2 - i (- 3 i)}{(2 + 3 i) (2 - 3 i)}) \cdot \frac{a}{b} = 1$

Step 1.1.5.2.2.1.2

Multiply $- 3$ by $5$ .

$(i + \frac{10 - 15 i - i \cdot 2 - i (- 3 i)}{(2 + 3 i) (2 - 3 i)}) \cdot \frac{a}{b} = 1$

Step 1.1.5.2.2.1.3

Multiply $2$ by $- 1$ .

$(i + \frac{10 - 15 i - 2 i - i (- 3 i)}{(2 + 3 i) (2 - 3 i)}) \cdot \frac{a}{b} = 1$

Step 1.1.5.2.2.1.4

Multiply $- i (- 3 i)$ .

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Step 1.1.5.2.2.1.4.1

Multiply $- 3$ by $- 1$ .

$(i + \frac{10 - 15 i - 2 i + 3 i i}{(2 + 3 i) (2 - 3 i)}) \cdot \frac{a}{b} = 1$

Step 1.1.5.2.2.1.4.2

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

$(i + \frac{10 - 15 i - 2 i + 3 (i^{1} i)}{(2 + 3 i) (2 - 3 i)}) \cdot \frac{a}{b} = 1$

Step 1.1.5.2.2.1.4.3

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

$(i + \frac{10 - 15 i - 2 i + 3 (i^{1} i^{1})}{(2 + 3 i) (2 - 3 i)}) \cdot \frac{a}{b} = 1$

Step 1.1.5.2.2.1.4.4

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

$(i + \frac{10 - 15 i - 2 i + 3 i^{1 + 1}}{(2 + 3 i) (2 - 3 i)}) \cdot \frac{a}{b} = 1$

Step 1.1.5.2.2.1.4.5

Add $1$ and $1$ .

$(i + \frac{10 - 15 i - 2 i + 3 i^{2}}{(2 + 3 i) (2 - 3 i)}) \cdot \frac{a}{b} = 1$

Step 1.1.5.2.2.1.5

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

$(i + \frac{10 - 15 i - 2 i + 3 \cdot - 1}{(2 + 3 i) (2 - 3 i)}) \cdot \frac{a}{b} = 1$

Step 1.1.5.2.2.1.6

Multiply $3$ by $- 1$ .

$(i + \frac{10 - 15 i - 2 i - 3}{(2 + 3 i) (2 - 3 i)}) \cdot \frac{a}{b} = 1$

Step 1.1.5.2.2.2

Subtract $3$ from $10$ .

$(i + \frac{7 - 15 i - 2 i}{(2 + 3 i) (2 - 3 i)}) \cdot \frac{a}{b} = 1$

Step 1.1.5.2.2.3

Subtract $2 i$ from $- 15 i$ .

$(i + \frac{7 - 17 i}{(2 + 3 i) (2 - 3 i)}) \cdot \frac{a}{b} = 1$

Step 1.1.5.3

Simplify the denominator.

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Step 1.1.5.3.1

Expand $(2 + 3 i) (2 - 3 i)$ using the FOIL Method.

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Step 1.1.5.3.1.1

Apply the distributive property.

$(i + \frac{7 - 17 i}{2 (2 - 3 i) + 3 i (2 - 3 i)}) \cdot \frac{a}{b} = 1$

Step 1.1.5.3.1.2

Apply the distributive property.

$(i + \frac{7 - 17 i}{2 \cdot 2 + 2 (- 3 i) + 3 i (2 - 3 i)}) \cdot \frac{a}{b} = 1$

Step 1.1.5.3.1.3

Apply the distributive property.

$(i + \frac{7 - 17 i}{2 \cdot 2 + 2 (- 3 i) + 3 i \cdot 2 + 3 i (- 3 i)}) \cdot \frac{a}{b} = 1$

Step 1.1.5.3.2

Simplify.

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Step 1.1.5.3.2.1

Multiply $2$ by $2$ .

$(i + \frac{7 - 17 i}{4 + 2 (- 3 i) + 3 i \cdot 2 + 3 i (- 3 i)}) \cdot \frac{a}{b} = 1$

Step 1.1.5.3.2.2

Multiply $- 3$ by $2$ .

$(i + \frac{7 - 17 i}{4 - 6 i + 3 i \cdot 2 + 3 i (- 3 i)}) \cdot \frac{a}{b} = 1$

Step 1.1.5.3.2.3

Multiply $2$ by $3$ .

$(i + \frac{7 - 17 i}{4 - 6 i + 6 i + 3 i (- 3 i)}) \cdot \frac{a}{b} = 1$

Step 1.1.5.3.2.4

Multiply $- 3$ by $3$ .

$(i + \frac{7 - 17 i}{4 - 6 i + 6 i - 9 i i}) \cdot \frac{a}{b} = 1$

Step 1.1.5.3.2.5

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

$(i + \frac{7 - 17 i}{4 - 6 i + 6 i - 9 (i^{1} i)}) \cdot \frac{a}{b} = 1$

Step 1.1.5.3.2.6

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

$(i + \frac{7 - 17 i}{4 - 6 i + 6 i - 9 (i^{1} i^{1})}) \cdot \frac{a}{b} = 1$

Step 1.1.5.3.2.7

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

$(i + \frac{7 - 17 i}{4 - 6 i + 6 i - 9 i^{1 + 1}}) \cdot \frac{a}{b} = 1$

Step 1.1.5.3.2.8

Add $1$ and $1$ .

$(i + \frac{7 - 17 i}{4 - 6 i + 6 i - 9 i^{2}}) \cdot \frac{a}{b} = 1$

Step 1.1.5.3.2.9

Add $- 6 i$ and $6 i$ .

$(i + \frac{7 - 17 i}{4 + 0 - 9 i^{2}}) \cdot \frac{a}{b} = 1$

Step 1.1.5.3.2.10

Add $4$ and $0$ .

$(i + \frac{7 - 17 i}{4 - 9 i^{2}}) \cdot \frac{a}{b} = 1$

Step 1.1.5.3.3

Simplify each term.

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Step 1.1.5.3.3.1

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

$(i + \frac{7 - 17 i}{4 - 9 \cdot - 1}) \cdot \frac{a}{b} = 1$

Step 1.1.5.3.3.2

Multiply $- 9$ by $- 1$ .

$(i + \frac{7 - 17 i}{4 + 9}) \cdot \frac{a}{b} = 1$

Step 1.1.5.3.4

Add $4$ and $9$ .

$(i + \frac{7 - 17 i}{13}) \cdot \frac{a}{b} = 1$

Step 1.1.6

Split the fraction $\frac{7 - 17 i}{13}$ into two fractions.

$(i + \frac{7}{13} + \frac{- 17 i}{13}) \cdot \frac{a}{b} = 1$

Step 1.1.7

Move the negative in front of the fraction.

$(i + \frac{7}{13} - \frac{17 i}{13}) \cdot \frac{a}{b} = 1$

Step 1.2

To write $i$ as a fraction with a common denominator, multiply by $\frac{13}{13}$ .

$(i \cdot \frac{13}{13} - \frac{17 i}{13} + \frac{7}{13}) \cdot \frac{a}{b} = 1$

Step 1.3

Combine fractions.

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Step 1.3.1

Combine $i$ and $\frac{13}{13}$ .

$(\frac{i \cdot 13}{13} - \frac{17 i}{13} + \frac{7}{13}) \cdot \frac{a}{b} = 1$

Step 1.3.2

Combine the numerators over the common denominator.

$(\frac{- 4 i}{13} + \frac{7}{13}) \cdot \frac{a}{b} = 1$

Step 1.4

Move the negative in front of the fraction.

$\frac{(- \frac{4 i}{13} + \frac{7}{13}) a}{b} = 1$

Step 1.5

Combine fractions.

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Step 1.5.1

Combine the numerators over the common denominator.

$\frac{\frac{7 - 4 i}{13} a}{b} = 1$

Step 1.5.2

Combine $\frac{7 - 4 i}{13}$ and $a$ .

$\frac{\frac{(7 - 4 i) a}{13}}{b} = 1$

Step 1.6

Multiply the numerator by the reciprocal of the denominator.

$\frac{(7 - 4 i) a}{13} \cdot \frac{1}{b} = 1$

Step 1.7

Multiply $\frac{(7 - 4 i) a}{13}$ by $\frac{1}{b}$ .

$\frac{(7 - 4 i) a}{13 b} = 1$

Step 1.8

Reorder factors in $\frac{(7 - 4 i) a}{13 b}$ .

$\frac{a (7 - 4 i)}{13 b} = 1$

Step 2

Find the LCD of the terms in the equation.

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Step 2.1

Finding the LCD of a list of values is the same as finding the LCM of the denominators of those values.

$13 b, 1$

Step 2.2

The LCM of one and any expression is the expression.

$13 b$

Step 3

Multiply each term in $\frac{a (7 - 4 i)}{13 b} = 1$ by $13 b$ to eliminate the fractions.

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Step 3.1

Multiply each term in $\frac{a (7 - 4 i)}{13 b} = 1$ by $13 b$ .

$\frac{a (7 - 4 i)}{13 b} (13 b) = 1 (13 b)$

Step 3.2

Simplify the left side.

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Step 3.2.1

Rewrite using the commutative property of multiplication.

$13 \frac{a (7 - 4 i)}{13 b} b = 1 (13 b)$

Step 3.2.2

Cancel the common factor of $13$ .

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Step 3.2.2.1

Factor $13$ out of $13 b$ .

$13 \frac{a (7 - 4 i)}{13 (b)} b = 1 (13 b)$

Step 3.2.2.2

Cancel the common factor.

$13 \frac{a (7 - 4 i)}{13 b} b = 1 (13 b)$

Step 3.2.2.3

Rewrite the expression.

$\frac{a (7 - 4 i)}{b} b = 1 (13 b)$

Step 3.2.3

Cancel the common factor of $b$ .

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Step 3.2.3.1

Cancel the common factor.

$\frac{a (7 - 4 i)}{b} b = 1 (13 b)$

Step 3.2.3.2

Rewrite the expression.

$a (7 - 4 i) = 1 (13 b)$

Step 3.2.4

Apply the distributive property.

$a \cdot 7 + a (- 4 i) = 1 (13 b)$

Step 3.2.5

Simplify the expression.

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Step 3.2.5.1

Move $7$ to the left of $a$ .

$7 \cdot a + a (- 4 i) = 1 (13 b)$

Step 3.2.5.2

Reorder factors in $7 a + a (- 4 i)$ .

$7 a - 4 a i = 1 (13 b)$

Step 3.3

Simplify the right side.

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Step 3.3.1

Multiply $13 b$ by $1$ .

$7 a - 4 a i = 13 b$

Step 4

Solve the equation.

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Step 4.1

Rewrite the equation as $13 b = 7 a - 4 a i$ .

$13 b = 7 a - 4 a i$

Step 4.2

Divide each term in $13 b = 7 a - 4 a i$ by $13$ and simplify.

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Step 4.2.1

Divide each term in $13 b = 7 a - 4 a i$ by $13$ .

$\frac{13 b}{13} = \frac{7 a}{13} + \frac{- 4 a i}{13}$

Step 4.2.2

Simplify the left side.

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Step 4.2.2.1

Cancel the common factor of $13$ .

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Step 4.2.2.1.1

Cancel the common factor.

$\frac{13 b}{13} = \frac{7 a}{13} + \frac{- 4 a i}{13}$

Step 4.2.2.1.2

Divide $b$ by $1$ .

$b = \frac{7 a}{13} + \frac{- 4 a i}{13}$

Step 4.2.3

Simplify the right side.

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Step 4.2.3.1

Move the negative in front of the fraction.

$b = \frac{7 a}{13} - \frac{4 a i}{13}$