Solve (2(1+2i)y+(6+4i)/2)-y+2yi=6+4i

$$( 2 ( 1 + 2 i ) \frac { y + ( 6 + 4 i ) } { 2 } ) - y + 2 y i = 6 + 4 i$$

Solve for y

$$y=-3-2i$$

Steps for Solving Linear Equation

Multiply both sides of the equation by 2.

$$2\left(1+2i\right)\left(y+\left(6+4i\right)\right)-2y+4yi=12+8i$$

Multiply 2 times 1+2i.

$$\left(2\times 1+2\times \left(2i\right)\right)\left(y+\left(6+4i\right)\right)-2y+4yi=12+8i$$

Do the multiplications in 2\times 1+2\times \left(2i\right).

$$\left(2+4i\right)\left(y+\left(6+4i\right)\right)-2y+4yi=12+8i$$

Use the distributive property to multiply 2+4i by y+\left(6+4i\right).

$$\left(2+4i\right)y+\left(-4+32i\right)-2y+4yi=12+8i$$

Combine \left(2+4i\right)y and -2y to get 4iy.

$$4iy+\left(-4+32i\right)+4yi=12+8i$$

Multiply 4 and i to get 4i.

$$4iy+\left(-4+32i\right)+4iy=12+8i$$

Combine 4iy and 4iy to get 8iy.

$$8iy+\left(-4+32i\right)=12+8i$$

Subtract -4+32i from both sides.

$$8iy=12+8i-\left(-4+32i\right)$$

Subtract -4+32i from 12+8i by subtracting corresponding real and imaginary parts.

$$8iy=12-\left(-4\right)+\left(8-32\right)i$$

Subtract -4 from 12. Subtract 32 from 8.

$$8iy=16-24i$$

Divide both sides by 8i.

$$y=\frac{16-24i}{8i}$$

Multiply both numerator and denominator of \frac{16-24i}{8i} by imaginary unit i.

$$y=\frac{\left(16-24i\right)i}{8i^{2}}$$

By definition, i^{2} is -1. Calculate the denominator.

$$y=\frac{\left(16-24i\right)i}{-8}$$

Multiply 16-24i times i.

$$y=\frac{16i-24i^{2}}{-8}$$

By definition, i^{2} is -1.

$$y=\frac{16i-24\left(-1\right)}{-8}$$

Do the multiplications in 16i-24\left(-1\right). Reorder the terms.

$$y=\frac{24+16i}{-8}$$

Divide 24+16i by -8 to get -3-2i.

$$y=-3-2i$$