Solve x/1-i+y/1-2i=5/1-3i

$$\frac { x } { 1 - i } + \frac { y } { 1 - 2 i } = \frac { 5 } { 1 - 3 i }$$

Solve for x

$$x=\left(-\frac{3}{5}-\frac{1}{5}i\right)y+\left(2+i\right)$$

Steps for Solving Linear Equation

Multiply both numerator and denominator of \frac{5}{1-3i} by the complex conjugate of the denominator, 1+3i.

$$\frac{x}{1-i}+\frac{y}{1-2i}=\frac{5\left(1+3i\right)}{\left(1-3i\right)\left(1+3i\right)}$$

Do the multiplications in \frac{5\left(1+3i\right)}{\left(1-3i\right)\left(1+3i\right)}.

$$\frac{x}{1-i}+\frac{y}{1-2i}=\frac{5+15i}{10}$$

Divide 5+15i by 10 to get \frac{1}{2}+\frac{3}{2}i.

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

Subtract \frac{y}{1-2i} from both sides.

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

The equation is in standard form.

$$\left(\frac{1}{2}+\frac{1}{2}i\right)x=\left(-\frac{1}{5}-\frac{2}{5}i\right)y+\left(\frac{1}{2}+\frac{3}{2}i\right)$$

Divide both sides by \frac{1}{2}+\frac{1}{2}i.

$$\frac{\left(\frac{1}{2}+\frac{1}{2}i\right)x}{\frac{1}{2}+\frac{1}{2}i}=\frac{\left(-\frac{1}{5}-\frac{2}{5}i\right)y+\left(\frac{1}{2}+\frac{3}{2}i\right)}{\frac{1}{2}+\frac{1}{2}i}$$

Dividing by \frac{1}{2}+\frac{1}{2}i undoes the multiplication by \frac{1}{2}+\frac{1}{2}i.

$$x=\frac{\left(-\frac{1}{5}-\frac{2}{5}i\right)y+\left(\frac{1}{2}+\frac{3}{2}i\right)}{\frac{1}{2}+\frac{1}{2}i}$$

Divide \frac{1}{2}+\frac{3}{2}i+\left(-\frac{1}{5}-\frac{2}{5}i\right)y by \frac{1}{2}+\frac{1}{2}i.

$$x=\left(-\frac{3}{5}-\frac{1}{5}i\right)y+\left(2+i\right)$$

Solve for y

$$y=\left(-\frac{3}{2}+\frac{1}{2}i\right)x+\left(\frac{7}{2}+\frac{1}{2}i\right)$$

Steps for Solving Linear Equation

Multiply both numerator and denominator of \frac{5}{1-3i} by the complex conjugate of the denominator, 1+3i.

$$\frac{x}{1-i}+\frac{y}{1-2i}=\frac{5\left(1+3i\right)}{\left(1-3i\right)\left(1+3i\right)}$$

Do the multiplications in \frac{5\left(1+3i\right)}{\left(1-3i\right)\left(1+3i\right)}.

$$\frac{x}{1-i}+\frac{y}{1-2i}=\frac{5+15i}{10}$$

Divide 5+15i by 10 to get \frac{1}{2}+\frac{3}{2}i.

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

Subtract \frac{x}{1-i} from both sides.

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

The equation is in standard form.

$$\left(\frac{1}{5}+\frac{2}{5}i\right)y=\left(-\frac{1}{2}-\frac{1}{2}i\right)x+\left(\frac{1}{2}+\frac{3}{2}i\right)$$

Divide both sides by \frac{1}{5}+\frac{2}{5}i.

$$\frac{\left(\frac{1}{5}+\frac{2}{5}i\right)y}{\frac{1}{5}+\frac{2}{5}i}=\frac{\left(-\frac{1}{2}-\frac{1}{2}i\right)x+\left(\frac{1}{2}+\frac{3}{2}i\right)}{\frac{1}{5}+\frac{2}{5}i}$$

Dividing by \frac{1}{5}+\frac{2}{5}i undoes the multiplication by \frac{1}{5}+\frac{2}{5}i.

$$y=\frac{\left(-\frac{1}{2}-\frac{1}{2}i\right)x+\left(\frac{1}{2}+\frac{3}{2}i\right)}{\frac{1}{5}+\frac{2}{5}i}$$

Divide \frac{1}{2}+\frac{3}{2}i+\left(-\frac{1}{2}-\frac{1}{2}i\right)x by \frac{1}{5}+\frac{2}{5}i.

$$y=\left(-\frac{3}{2}+\frac{1}{2}i\right)x+\left(\frac{7}{2}+\frac{1}{2}i\right)$$