Solve in Terms of the Arbitrary Variable y ((5-3i)(x+iy))/(4-5i)=(2+i)^2-(3-4i)^2

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

Step 1

Solve the equation for $i$ .

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

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

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

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

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

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

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

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

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

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

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

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

Simplify each term.

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

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

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

Apply the distributive property.

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

Apply the distributive property.

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

Apply the distributive property.

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

Simplify and combine like terms.

Simplify each term.

Multiply $2$ by $2$ .

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

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

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

Multiply $i$ by $i$ .

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

Add $2 i$ and $2 i$ .

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

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

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

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

Apply the distributive property.

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

Apply the distributive property.

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

Apply the distributive property.

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

Simplify and combine like terms.

Simplify each term.

Multiply $3$ by $3$ .

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

Multiply $- 4$ by $3$ .

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

Multiply $3$ by $- 4$ .

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

Rewrite using the commutative property of multiplication.

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

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

Move $i$ .

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

Multiply $i$ by $i$ .

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

Multiply $- 4$ by $- 4$ .

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

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

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

Apply the distributive property.

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

Simplify.

Multiply $- 1$ by $9$ .

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

Multiply $- 24$ by $- 1$ .

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

Multiply $16$ by $- 1$ .

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

Simplify by adding terms.

Subtract $9$ from $4$ .

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

Add $4 i$ and $24 i$ .

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

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

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

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

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

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

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

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

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

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

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

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

Combine the numerators over the common denominator.

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

Simplify each term.

Simplify the numerator.

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

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

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

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

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

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

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

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

Apply the distributive property.

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

Apply the distributive property.

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

Apply the distributive property.

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

Simplify each term.

Multiply $5$ by $1$ .

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

Multiply $- 3$ by $1$ .

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

Apply the distributive property.

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

Multiply $- 28$ by $4$ .

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

Multiply $- 5$ by $- 28$ .

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

Subtract $112$ from $5$ .

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

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

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

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

Regroup terms.

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

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

Reorder the expression.

Move $5 y$ .

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

Factor out negative.

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

Move the negative in front of the fraction.

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

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

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

Combine the numerators over the common denominator.

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

Simplify the numerator.

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

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

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

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

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

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

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

Apply the distributive property.

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

Simplify.

Multiply $3$ by $- 1$ .

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

Multiply $- 137$ by $- 1$ .

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

Multiply $- 5$ by $- 1$ .

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

Multiply $- 1$ by $107$ .

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

Apply the distributive property.

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

Multiply $4$ by $15$ .

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

Rewrite using the commutative property of multiplication.

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

Simplify each term.

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

Move $i$ .

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

Multiply $i$ by $i$ .

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

Multiply $15$ by $- 5$ .

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

Add $137 i$ and $60 i$ .

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

Move the negative in front of the fraction.

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

Step 2

Solve the equation for $x$ .

Multiply both sides by $4 - 5 \cdot - 5$ .

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

Simplify.

Simplify the left side.

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

Reduce the expression by cancelling the common factors.

Cancel the common factor of $4 - 5 \cdot - 5$ .

Cancel the common factor.

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

Rewrite the expression.

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

Simplify the expression.

Multiply $- 3$ by $- 5$ .

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

Add $5$ and $15$ .

$20 (x + (- 5) \cdot y) = ({(2 - 5)}^{2} - {(3 - 4 \cdot - 5)}^{2}) (4 - 5 \cdot - 5)$

Multiply $- 5$ by $y$ .

$20 (x - 5 y) = ({(2 - 5)}^{2} - {(3 - 4 \cdot - 5)}^{2}) (4 - 5 \cdot - 5)$

Simplify by multiplying through.

Apply the distributive property.

$20 x + 20 (- 5 y) = ({(2 - 5)}^{2} - {(3 - 4 \cdot - 5)}^{2}) (4 - 5 \cdot - 5)$

Multiply $- 5$ by $20$ .

$20 x - 100 y = ({(2 - 5)}^{2} - {(3 - 4 \cdot - 5)}^{2}) (4 - 5 \cdot - 5)$

Simplify the right side.

Simplify $({(2 - 5)}^{2} - {(3 - 4 \cdot - 5)}^{2}) (4 - 5 \cdot - 5)$ .

Simplify each term.

Subtract $5$ from $2$ .

$20 x - 100 y = ({(- 3)}^{2} - {(3 - 4 \cdot - 5)}^{2}) (4 - 5 \cdot - 5)$

Raise $- 3$ to the power of $2$ .

$20 x - 100 y = (9 - {(3 - 4 \cdot - 5)}^{2}) (4 - 5 \cdot - 5)$

Multiply $- 4$ by $- 5$ .

$20 x - 100 y = (9 - {(3 + 20)}^{2}) (4 - 5 \cdot - 5)$

Add $3$ and $20$ .

$20 x - 100 y = (9 - 23^{2}) (4 - 5 \cdot - 5)$

Raise $23$ to the power of $2$ .

$20 x - 100 y = (9 - 1 \cdot 529) (4 - 5 \cdot - 5)$

Multiply $- 1$ by $529$ .

$20 x - 100 y = (9 - 529) (4 - 5 \cdot - 5)$

Simplify the expression.

Subtract $529$ from $9$ .

$20 x - 100 y = - 520 (4 - 5 \cdot - 5)$

Multiply $- 5$ by $- 5$ .

$20 x - 100 y = - 520 (4 + 25)$

Add $4$ and $25$ .

$20 x - 100 y = - 520 \cdot 29$

Multiply $- 520$ by $29$ .

$20 x - 100 y = - 15080$

Solve for $x$ .

Add $100 y$ to both sides of the equation.

$20 x = - 15080 + 100 y$

Divide each term in $20 x = - 15080 + 100 y$ by $20$ and simplify.

Divide each term in $20 x = - 15080 + 100 y$ by $20$ .

$\frac{20 x}{20} = \frac{- 15080}{20} + \frac{100 y}{20}$

Simplify the left side.

Cancel the common factor of $20$ .

Cancel the common factor.

$\frac{20 x}{20} = \frac{- 15080}{20} + \frac{100 y}{20}$

Divide $x$ by $1$ .

$x = \frac{- 15080}{20} + \frac{100 y}{20}$

Simplify the right side.

Simplify each term.

Divide $- 15080$ by $20$ .

$x = - 754 + \frac{100 y}{20}$

Cancel the common factor of $100$ and $20$ .

Factor $20$ out of $100 y$ .

$x = - 754 + \frac{20 (5 y)}{20}$

Cancel the common factors.

Factor $20$ out of $20$ .

$x = - 754 + \frac{20 (5 y)}{20 (1)}$

Cancel the common factor.

$x = - 754 + \frac{20 (5 y)}{20 \cdot 1}$

Rewrite the expression.

$x = - 754 + \frac{5 y}{1}$

Divide $5 y$ by $1$ .

$x = - 754 + 5 y$

Step 3

Simplify the right side.

Reorder $- 754$ and $5 y$ .

$x = 5 y - 754$

Solve in Terms of the Arbitrary Variable y ((5-3i)(x+iy))/(4-5i)=(2+i)^2-(3-4i)^2

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