Solve (x+y)-(-3x+y)i=2+i/1-i

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

Solve for x

$$x=\left(\frac{1}{5}+\frac{2}{5}i\right)y+\frac{1}{2}$$

Steps for Solving Linear Equation

Use the distributive property to multiply -3x+y by i.

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

To find the opposite of -3ix+iy, find the opposite of each term.

$$x+y+3ix-iy=\frac{2+i}{1-i}$$

Combine x and 3ix to get \left(1+3i\right)x.

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

Combine y and -iy to get \left(1-i\right)y.

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

Multiply both numerator and denominator of \frac{2+i}{1-i} by the complex conjugate of the denominator, 1+i.

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

Do the multiplications in \frac{\left(2+i\right)\left(1+i\right)}{\left(1-i\right)\left(1+i\right)}.

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

Divide 1+3i by 2 to get \frac{1}{2}+\frac{3}{2}i.

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

Subtract \left(1-i\right)y from both sides.

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

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

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

The equation is in standard form.

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

Divide both sides by 1+3i.

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

Dividing by 1+3i undoes the multiplication by 1+3i.

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

Divide \frac{1}{2}+\frac{3}{2}i+\left(-1+i\right)y by 1+3i.

$$x=\left(\frac{1}{5}+\frac{2}{5}i\right)y+\frac{1}{2}$$

Solve for y

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

Steps for Solving Linear Equation

Use the distributive property to multiply -3x+y by i.

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

To find the opposite of -3ix+iy, find the opposite of each term.

$$x+y+3ix-iy=\frac{2+i}{1-i}$$

Combine x and 3ix to get \left(1+3i\right)x.

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

Combine y and -iy to get \left(1-i\right)y.

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

Multiply both numerator and denominator of \frac{2+i}{1-i} by the complex conjugate of the denominator, 1+i.

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

Do the multiplications in \frac{\left(2+i\right)\left(1+i\right)}{\left(1-i\right)\left(1+i\right)}.

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

Divide 1+3i by 2 to get \frac{1}{2}+\frac{3}{2}i.

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

Subtract \left(1+3i\right)x from both sides.

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

Multiply -1 and 1+3i to get -1-3i.

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

The equation is in standard form.

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

Divide both sides by 1-i.

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

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

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

Divide \frac{1}{2}+\frac{3}{2}i+\left(-1-3i\right)x by 1-i.

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