Algebra Examples

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

Step 1

Simplify $\frac{(5 - 3 i) (i + i \cdot y)}{4 - 5 i}$ .

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Step 1.1

Factor $i$ out of $i + i \cdot y$ .

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Step 1.1.1

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

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

Step 1.1.2

Factor $i$ out of $i^{1}$ .

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

Step 1.1.3

Factor $i$ out of $i \cdot 1 + i y$ .

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

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

Step 1.2

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

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

Step 2

Simplify ${(2 + i)}^{2} - {(3 - 4 i)}^{2}$ .

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Step 2.1

Simplify each term.

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Step 2.1.1

Rewrite ${(2 + i)}^{2}$ as $(2 + i) (2 + i)$ .

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

Step 2.1.2

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

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Step 2.1.2.1

Apply the distributive property.

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

Step 2.1.2.2

Apply the distributive property.

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

Step 2.1.2.3

Apply the distributive property.

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

Step 2.1.3

Simplify and combine like terms.

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Step 2.1.3.1

Simplify each term.

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Step 2.1.3.1.1

Multiply $2$ by $2$ .

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

Step 2.1.3.1.2

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

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

Step 2.1.3.1.3

Multiply $i$ by $i$ .

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

Step 2.1.3.2

Add $2 i$ and $2 i$ .

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

Step 2.1.4

Rewrite ${(3 - 4 i)}^{2}$ as $(3 - 4 i) (3 - 4 i)$ .

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

Step 2.1.5

Expand $(3 - 4 i) (3 - 4 i)$ using the FOIL Method.

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Step 2.1.5.1

Apply the distributive property.

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

Step 2.1.5.2

Apply the distributive property.

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

Step 2.1.5.3

Apply the distributive property.

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

Step 2.1.6

Simplify and combine like terms.

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Step 2.1.6.1

Simplify each term.

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Step 2.1.6.1.1

Multiply $3$ by $3$ .

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

Step 2.1.6.1.2

Multiply $- 4$ by $3$ .

$\frac{i (5 - 3 i) (1 + y)}{4 - 5 i} = 4 + 4 i + i^{2} - (9 - 12 i - 4 i \cdot 3 - 4 i (- 4 i))$

Step 2.1.6.1.3

Multiply $3$ by $- 4$ .

$\frac{i (5 - 3 i) (1 + y)}{4 - 5 i} = 4 + 4 i + i^{2} - (9 - 12 i - 12 i - 4 i (- 4 i))$

Step 2.1.6.1.4

Rewrite using the commutative property of multiplication.

$\frac{i (5 - 3 i) (1 + y)}{4 - 5 i} = 4 + 4 i + i^{2} - (9 - 12 i - 12 i - 4 \cdot - 4 i i)$

Step 2.1.6.1.5

Multiply $i$ by $i$ by adding the exponents.

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Step 2.1.6.1.5.1

Move $i$ .

$\frac{i (5 - 3 i) (1 + y)}{4 - 5 i} = 4 + 4 i + i^{2} - (9 - 12 i - 12 i - 4 \cdot - 4 (i i))$

Step 2.1.6.1.5.2

Multiply $i$ by $i$ .

$\frac{i (5 - 3 i) (1 + y)}{4 - 5 i} = 4 + 4 i + i^{2} - (9 - 12 i - 12 i - 4 \cdot - 4 i^{2})$

Step 2.1.6.1.6

Multiply $- 4$ by $- 4$ .

$\frac{i (5 - 3 i) (1 + y)}{4 - 5 i} = 4 + 4 i + i^{2} - (9 - 12 i - 12 i + 16 i^{2})$

Step 2.1.6.2

Subtract $12 i$ from $- 12 i$ .

$\frac{i (5 - 3 i) (1 + y)}{4 - 5 i} = 4 + 4 i + i^{2} - (9 - 24 i + 16 i^{2})$

Step 2.1.7

Apply the distributive property.

$\frac{i (5 - 3 i) (1 + y)}{4 - 5 i} = 4 + 4 i + i^{2} - 1 \cdot 9 - (- 24 i) - (16 i^{2})$

Step 2.1.8

Simplify.

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Step 2.1.8.1

Multiply $- 1$ by $9$ .

$\frac{i (5 - 3 i) (1 + y)}{4 - 5 i} = 4 + 4 i + i^{2} - 9 - (- 24 i) - (16 i^{2})$

Step 2.1.8.2

Multiply $- 24$ by $- 1$ .

$\frac{i (5 - 3 i) (1 + y)}{4 - 5 i} = 4 + 4 i + i^{2} - 9 + 24 i - (16 i^{2})$

Step 2.1.8.3

Multiply $16$ by $- 1$ .

$\frac{i (5 - 3 i) (1 + y)}{4 - 5 i} = 4 + 4 i + i^{2} - 9 + 24 i - 16 i^{2}$

Step 2.2

Simplify by adding terms.

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Step 2.2.1

Subtract $9$ from $4$ .

$\frac{i (5 - 3 i) (1 + y)}{4 - 5 i} = 4 i + i^{2} - 5 + 24 i - 16 i^{2}$

Step 2.2.2

Add $4 i$ and $24 i$ .

$\frac{i (5 - 3 i) (1 + y)}{4 - 5 i} = 28 i + i^{2} - 5 - 16 i^{2}$

Step 2.2.3

Subtract $16 i^{2}$ from $i^{2}$ .

$\frac{i (5 - 3 i) (1 + y)}{4 - 5 i} = 28 i - 15 i^{2} - 5$

Step 3

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

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Step 3.1

Subtract $28 i$ from both sides of the equation.

$\frac{i (5 - 3 i) (1 + y)}{4 - 5 i} - 28 i = - 15 i^{2} - 5$

Step 3.2

Add $15 i^{2}$ to both sides of the equation.

$\frac{i (5 - 3 i) (1 + y)}{4 - 5 i} - 28 i + 15 i^{2} = - 5$

Step 3.3

To write $- 28 i$ as a fraction with a common denominator, multiply by $\frac{4 - 5 i}{4 - 5 i}$ .

$\frac{i (5 - 3 i) (1 + y)}{4 - 5 i} - 28 i \cdot \frac{4 - 5 i}{4 - 5 i} + 15 i^{2} = - 5$

Step 3.4

Combine $- 28 i$ and $\frac{4 - 5 i}{4 - 5 i}$ .

$\frac{i (5 - 3 i) (1 + y)}{4 - 5 i} + \frac{- 28 i (4 - 5 i)}{4 - 5 i} + 15 i^{2} = - 5$

Step 3.5

Combine the numerators over the common denominator.

$\frac{i (5 - 3 i) (1 + y) - 28 i (4 - 5 i)}{4 - 5 i} + 15 i^{2} = - 5$

Step 3.6

Simplify each term.

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Step 3.6.1

Simplify the numerator.

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Step 3.6.1.1

Factor $i$ out of $i (5 - 3 i) (1 + y) - 28 i (4 - 5 i)$ .

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Step 3.6.1.1.1

Factor $i$ out of $i (5 - 3 i) (1 + y)$ .

$\frac{i ((5 - 3 i) (1 + y)) - 28 i (4 - 5 i)}{4 - 5 i} + 15 i^{2} = - 5$

Step 3.6.1.1.2

Factor $i$ out of $- 28 i (4 - 5 i)$ .

$\frac{i ((5 - 3 i) (1 + y)) + i (- 28 (4 - 5 i))}{4 - 5 i} + 15 i^{2} = - 5$

Step 3.6.1.1.3

Factor $i$ out of $i ((5 - 3 i) (1 + y)) + i (- 28 (4 - 5 i))$ .

$\frac{i ((5 - 3 i) (1 + y) - 28 (4 - 5 i))}{4 - 5 i} + 15 i^{2} = - 5$

Step 3.6.1.2

Expand $(5 - 3 i) (1 + y)$ using the FOIL Method.

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Step 3.6.1.2.1

Apply the distributive property.

$\frac{i (5 (1 + y) - 3 i (1 + y) - 28 (4 - 5 i))}{4 - 5 i} + 15 i^{2} = - 5$

Step 3.6.1.2.2

Apply the distributive property.

$\frac{i (5 \cdot 1 + 5 y - 3 i (1 + y) - 28 (4 - 5 i))}{4 - 5 i} + 15 i^{2} = - 5$

Step 3.6.1.2.3

Apply the distributive property.

$\frac{i (5 \cdot 1 + 5 y - 3 i \cdot 1 - 3 i y - 28 (4 - 5 i))}{4 - 5 i} + 15 i^{2} = - 5$

Step 3.6.1.3

Simplify each term.

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Step 3.6.1.3.1

Multiply $5$ by $1$ .

$\frac{i (5 + 5 y - 3 i \cdot 1 - 3 i y - 28 (4 - 5 i))}{4 - 5 i} + 15 i^{2} = - 5$

Step 3.6.1.3.2

Multiply $- 3$ by $1$ .

$\frac{i (5 + 5 y - 3 i - 3 i y - 28 (4 - 5 i))}{4 - 5 i} + 15 i^{2} = - 5$

Step 3.6.1.4

Apply the distributive property.

$\frac{i (5 + 5 y - 3 i - 3 i y - 28 \cdot 4 - 28 (- 5 i))}{4 - 5 i} + 15 i^{2} = - 5$

Step 3.6.1.5

Multiply $- 28$ by $4$ .

$\frac{i (5 + 5 y - 3 i - 3 i y - 112 - 28 (- 5 i))}{4 - 5 i} + 15 i^{2} = - 5$

Step 3.6.1.6

Multiply $- 5$ by $- 28$ .

$\frac{i (5 + 5 y - 3 i - 3 i y - 112 + 140 i)}{4 - 5 i} + 15 i^{2} = - 5$

Step 3.6.1.7

Subtract $112$ from $5$ .

$\frac{i (5 y - 3 i - 3 i y - 107 + 140 i)}{4 - 5 i} + 15 i^{2} = - 5$

Step 3.6.1.8

Add $- 3 i$ and $140 i$ .

$\frac{i (5 y + 137 i - 3 i y - 107)}{4 - 5 i} + 15 i^{2} = - 5$

Step 3.6.1.9

Rewrite $5 y + 137 i - 3 i y - 107$ in a factored form.

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Step 3.6.1.9.1

Regroup terms.

$\frac{i (- 107 + 5 y + 137 i - 3 i y)}{4 - 5 i} + 15 i^{2} = - 5$

Step 3.6.1.9.2

Factor $- 1$ out of $5 y + 137 i - 3 i y$ .

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Step 3.6.1.9.2.1

Reorder the expression.

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Step 3.6.1.9.2.1.1

Move $5 y$ .

$\frac{i (- 107 + 137 i - 3 i y + 5 y)}{4 - 5 i} + 15 i^{2} = - 5$

Step 3.6.1.9.2.1.2

Reorder $137 i$ and $- 3 i y$ .

$\frac{i (- 107 - 3 i y + 137 i + 5 y)}{4 - 5 i} + 15 i^{2} = - 5$

Step 3.6.1.9.2.2

Factor $- 1$ out of $- 3 i y$ .

$\frac{i (- 107 - (3 i y) + 137 i + 5 y)}{4 - 5 i} + 15 i^{2} = - 5$

Step 3.6.1.9.2.3

Factor $- 1$ out of $137 i$ .

$\frac{i (- 107 - (3 i y) - (- 137 i) + 5 y)}{4 - 5 i} + 15 i^{2} = - 5$

Step 3.6.1.9.2.4

Factor $- 1$ out of $5 y$ .

$\frac{i (- 107 - (3 i y) - (- 137 i) - (- 5 y))}{4 - 5 i} + 15 i^{2} = - 5$

Step 3.6.1.9.2.5

Factor $- 1$ out of $- (3 i y) - (- 137 i)$ .

$\frac{i (- 107 - (3 i y - 137 i) - (- 5 y))}{4 - 5 i} + 15 i^{2} = - 5$

Step 3.6.1.9.2.6

Factor $- 1$ out of $- (3 i y - 137 i) - (- 5 y)$ .

$\frac{i (- 107 - (3 i y - 137 i - 5 y))}{4 - 5 i} + 15 i^{2} = - 5$

Step 3.6.1.9.3

Factor $- 1$ out of $- 107 - (3 i y - 137 i - 5 y)$ .

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Step 3.6.1.9.3.1

Reorder $- 107$ and $- (3 i y - 137 i - 5 y)$ .

$\frac{i (- (3 i y - 137 i - 5 y) - 107)}{4 - 5 i} + 15 i^{2} = - 5$

Step 3.6.1.9.3.2

Rewrite $- 107$ as $- 1 (107)$ .

$\frac{i (- (3 i y - 137 i - 5 y) - 1 (107))}{4 - 5 i} + 15 i^{2} = - 5$

Step 3.6.1.9.3.3

Factor $- 1$ out of $- (3 i y - 137 i - 5 y) - 1 (107)$ .

$\frac{i (- (3 i y - 137 i - 5 y + 107))}{4 - 5 i} + 15 i^{2} = - 5$

$\frac{i \cdot - 1 (3 i y - 137 i - 5 y + 107)}{4 - 5 i} + 15 i^{2} = - 5$

Step 3.6.1.10

Factor out negative.

$\frac{- i (3 i y - 137 i - 5 y + 107)}{4 - 5 i} + 15 i^{2} = - 5$

Step 3.6.2

Move the negative in front of the fraction.

$- \frac{i (3 i y - 137 i - 5 y + 107)}{4 - 5 i} + 15 i^{2} = - 5$

Step 3.7

To write $15 i^{2}$ as a fraction with a common denominator, multiply by $\frac{4 - 5 i}{4 - 5 i}$ .

$- \frac{i (3 i y - 137 i - 5 y + 107)}{4 - 5 i} + \frac{15 i^{2} (4 - 5 i)}{4 - 5 i} = - 5$

Step 3.8

Combine the numerators over the common denominator.

$\frac{- i (3 i y - 137 i - 5 y + 107) + 15 i^{2} (4 - 5 i)}{4 - 5 i} = - 5$

Step 3.9

Simplify the numerator.

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Step 3.9.1

Factor $i$ out of $- i (3 i y - 137 i - 5 y + 107) + 15 i^{2} (4 - 5 i)$ .

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Step 3.9.1.1

Factor $i$ out of $- i (3 i y - 137 i - 5 y + 107)$ .

$\frac{i (- (3 i y - 137 i - 5 y + 107)) + 15 i^{2} (4 - 5 i)}{4 - 5 i} = - 5$

Step 3.9.1.2

Factor $i$ out of $15 i^{2} (4 - 5 i)$ .

$\frac{i (- (3 i y - 137 i - 5 y + 107)) + i (15 i (4 - 5 i))}{4 - 5 i} = - 5$

Step 3.9.1.3

Factor $i$ out of $i (- (3 i y - 137 i - 5 y + 107)) + i (15 i (4 - 5 i))$ .

$\frac{i (- (3 i y - 137 i - 5 y + 107) + 15 i (4 - 5 i))}{4 - 5 i} = - 5$

Step 3.9.2

Apply the distributive property.

$\frac{i (- (3 i y) - (- 137 i) - (- 5 y) - 1 \cdot 107 + 15 i (4 - 5 i))}{4 - 5 i} = - 5$

Step 3.9.3

Simplify.

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Step 3.9.3.1

Multiply $3$ by $- 1$ .

$\frac{i (- 3 (i y) - (- 137 i) - (- 5 y) - 1 \cdot 107 + 15 i (4 - 5 i))}{4 - 5 i} = - 5$

Step 3.9.3.2

Multiply $- 137$ by $- 1$ .

$\frac{i (- 3 (i y) + 137 i - (- 5 y) - 1 \cdot 107 + 15 i (4 - 5 i))}{4 - 5 i} = - 5$

Step 3.9.3.3

Multiply $- 5$ by $- 1$ .

$\frac{i (- 3 (i y) + 137 i + 5 y - 1 \cdot 107 + 15 i (4 - 5 i))}{4 - 5 i} = - 5$

Step 3.9.3.4

Multiply $- 1$ by $107$ .

$\frac{i (- 3 (i y) + 137 i + 5 y - 107 + 15 i (4 - 5 i))}{4 - 5 i} = - 5$

Step 3.9.4

Apply the distributive property.

$\frac{i (- 3 i y + 137 i + 5 y - 107 + 15 i \cdot 4 + 15 i (- 5 i))}{4 - 5 i} = - 5$

Step 3.9.5

Multiply $4$ by $15$ .

$\frac{i (- 3 i y + 137 i + 5 y - 107 + 60 i + 15 i (- 5 i))}{4 - 5 i} = - 5$

Step 3.9.6

Rewrite using the commutative property of multiplication.

$\frac{i (- 3 i y + 137 i + 5 y - 107 + 60 i + 15 \cdot - 5 i i)}{4 - 5 i} = - 5$

Step 3.9.7

Simplify each term.

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Step 3.9.7.1

Multiply $i$ by $i$ by adding the exponents.

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Step 3.9.7.1.1

Move $i$ .

$\frac{i (- 3 i y + 137 i + 5 y - 107 + 60 i + 15 \cdot - 5 (i i))}{4 - 5 i} = - 5$

Step 3.9.7.1.2

Multiply $i$ by $i$ .

$\frac{i (- 3 i y + 137 i + 5 y - 107 + 60 i + 15 \cdot - 5 i^{2})}{4 - 5 i} = - 5$

Step 3.9.7.2

Multiply $15$ by $- 5$ .

$\frac{i (- 3 i y + 137 i + 5 y - 107 + 60 i - 75 i^{2})}{4 - 5 i} = - 5$

Step 3.9.8

Add $137 i$ and $60 i$ .

$\frac{i (- 3 i y + 197 i + 5 y - 107 - 75 i^{2})}{4 - 5 i} = - 5$

Step 3.10

Factor $- 1$ out of $- 3 i y$ .

$\frac{i (- (3 i y) + 197 i + 5 y - 107 - 75 i^{2})}{4 - 5 i} = - 5$

Step 3.11

Factor $- 1$ out of $197 i$ .

$\frac{i (- (3 i y) - (- 197 i) + 5 y - 107 - 75 i^{2})}{4 - 5 i} = - 5$

Step 3.12

Factor $- 1$ out of $- (3 i y) - (- 197 i)$ .

$\frac{i (- (3 i y - 197 i) + 5 y - 107 - 75 i^{2})}{4 - 5 i} = - 5$

Step 3.13

Factor $- 1$ out of $5 y$ .

$\frac{i (- (3 i y - 197 i) - (- 5 y) - 107 - 75 i^{2})}{4 - 5 i} = - 5$

Step 3.14

Factor $- 1$ out of $- (3 i y - 197 i) - (- 5 y)$ .

$\frac{i (- (3 i y - 197 i - 5 y) - 107 - 75 i^{2})}{4 - 5 i} = - 5$

Step 3.15

Rewrite $- 107$ as $- 1 (107)$ .

$\frac{i (- (3 i y - 197 i - 5 y) - 1 (107) - 75 i^{2})}{4 - 5 i} = - 5$

Step 3.16

Factor $- 1$ out of $- (3 i y - 197 i - 5 y) - 1 (107)$ .

$\frac{i (- (3 i y - 197 i - 5 y + 107) - 75 i^{2})}{4 - 5 i} = - 5$

Step 3.17

Factor $- 1$ out of $- 75 i^{2}$ .

$\frac{i (- (3 i y - 197 i - 5 y + 107) - (75 i^{2}))}{4 - 5 i} = - 5$

Step 3.18

Factor $- 1$ out of $- (3 i y - 197 i - 5 y + 107) - (75 i^{2})$ .

$\frac{i (- (3 i y - 197 i - 5 y + 107 + 75 i^{2}))}{4 - 5 i} = - 5$

Step 3.19

Rewrite $- (3 i y - 197 i - 5 y + 107 + 75 i^{2})$ as $- 1 (3 i y - 197 i - 5 y + 107 + 75 i^{2})$ .

$\frac{i (- 1 (3 i y - 197 i - 5 y + 107 + 75 i^{2}))}{4 - 5 i} = - 5$

Step 3.20

Move the negative in front of the fraction.

$- \frac{i (3 i y - 197 i - 5 y + 107 + 75 i^{2})}{4 - 5 i} = - 5$