Solve x+2y+left(x-yright)i=3-6texttt{i}

$$x+2y+ \left( x-y \right) i = 3-6 \texttt{i}$$

Solve for x

$$x=\left(-\frac{1}{2}+\frac{3}{2}i\right)y+\left(-\frac{3}{2}-\frac{9}{2}i\right)$$

Steps for Solving Linear Equation

Use the distributive property to multiply x-y by i.

$$x+2y+ix-iy=3-6i$$

Combine x and ix to get \left(1+i\right)x.

$$\left(1+i\right)x+2y-iy=3-6i$$

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

$$\left(1+i\right)x+\left(2-i\right)y=3-6i$$

Subtract \left(2-i\right)y from both sides.

$$\left(1+i\right)x=3-6i-\left(2-i\right)y$$

Multiply -1 and 2-i to get -2+i.

$$\left(1+i\right)x=3-6i+\left(-2+i\right)y$$

The equation is in standard form.

$$\left(1+i\right)x=\left(-2+i\right)y+\left(3-6i\right)$$

Divide both sides by 1+i.

$$\frac{\left(1+i\right)x}{1+i}=\frac{\left(-2+i\right)y+\left(3-6i\right)}{1+i}$$

Dividing by 1+i undoes the multiplication by 1+i.

$$x=\frac{\left(-2+i\right)y+\left(3-6i\right)}{1+i}$$

Divide 3-6i+\left(-2+i\right)y by 1+i.

$$x=\left(-\frac{1}{2}+\frac{3}{2}i\right)y+\left(-\frac{3}{2}-\frac{9}{2}i\right)$$

Solve for y

$$y=\left(-\frac{1}{5}-\frac{3}{5}i\right)x+\left(\frac{12}{5}-\frac{9}{5}i\right)$$

Steps for Solving Linear Equation

Use the distributive property to multiply x-y by i.

$$x+2y+ix-iy=3-6i$$

Combine x and ix to get \left(1+i\right)x.

$$\left(1+i\right)x+2y-iy=3-6i$$

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

$$\left(1+i\right)x+\left(2-i\right)y=3-6i$$

Subtract \left(1+i\right)x from both sides.

$$\left(2-i\right)y=3-6i-\left(1+i\right)x$$

Multiply -1 and 1+i to get -1-i.

$$\left(2-i\right)y=3-6i+\left(-1-i\right)x$$

The equation is in standard form.

$$\left(2-i\right)y=\left(-1-i\right)x+\left(3-6i\right)$$

Divide both sides by 2-i.

$$\frac{\left(2-i\right)y}{2-i}=\frac{\left(-1-i\right)x+\left(3-6i\right)}{2-i}$$

Dividing by 2-i undoes the multiplication by 2-i.

$$y=\frac{\left(-1-i\right)x+\left(3-6i\right)}{2-i}$$

Divide 3-6i+\left(-1-i\right)x by 2-i.

$$y=\left(-\frac{1}{5}-\frac{3}{5}i\right)x+\left(\frac{12}{5}-\frac{9}{5}i\right)$$